How to calculate voltage divider transistor bias?
Usually, you know what \$V_\text{CC}\$ is and you do in this case. It's \$V_\text{CC}=10\:\text{V}\$.

You also usually decide on the quiescent current, \$I_{\text{C}_Q}\$, for the CE amplifier as a matter of knowing what you have to drive with it (it rarely stands alone) and knowing something from the datasheet for the BJT. For example, let's look at this datasheet: PN2222A.



This sets a limit based on the worst case \$V_\text{CE}\$, which in this case might be perhaps a couple of volts less than our \$V_\text{CC}\$. If we want to limit the temperature rise of the BJT to a worst case of \$30^\circ\text{C}\$, then this means \$I_{\text{C}_\text{MAX}}\approx 18 \:\text{mA}\$. Looking at the following curves:



We can see a very convenient (and reasonable, perhaps) curve at \$I_{\text{C}_Q}=10\:\text{mA}\$. Personally, I think this is the maximum I'd accept for this device. It's enough underneath the temperature rise limitations to make me feel better and it's got a nice curve behavior in the chart, too. So let's call it that way.

(The upshot of this is that I cannot get to \$I_{\text{C}_Q}=100\:\text{mA}\$ with the PN2222A in a TO-92 package. However, you are perfectly free to select a different device with a different datasheet and curves and capabilities, if you feel the need for a quiescent current as high as you indicated.)

I generally want the DC point for the emitter to be at least \$1\:\text{V}\$ above ground, to aid temperature stability due to the emitter's AC \$r_e=\frac{k\cdot T}{q\cdot I_\text{C}}\$ dependence on T. More is better for this purpose. But more also cuts into the available collector voltage swing.

This is where some judgment comes into play. Requiring a lot of collector swing also implies a lot of variation in the collector current itself. And this impacts \$r_e\$, too. (Not just temperature, but also the collector current.) Here again, the more voltage you can afford to give to the DC quiescent point of the emitter, the better. Yes, it cuts into the collector swing... but this also means less collector current variation and therefore... more stability.

You've said nothing about the load that has to be driven here. So I'm just going to assume that there is no load, at all. (Or, not much.) So we'll just design it for the maximum swing possible given some other decisions that we've yet to conclude.

In addition to the DC emitter voltage value, there is also the minimum \$V_\text{CE}\$ for the BJT. I like to see at least \$2\:\text{V}\$ reserved here (partly because of the Early Effect and partly because it keeps the device well out of saturation.) However, how much you are willing to sacrifice depends also on what your \$V_\text{CE}\$ happens to be. I think I'll stick with \$V_{\text{CE}_\text{MIN}}=2\:\text{V}\$.

Now I can work out that the minimum collector voltage is \$V_{\text{C}_\text{MIN}}=1\:\text{V}+2\:\text{V}=3\:\text{V}\$. This means I have a total of \$V_\text{CC}-V_{\text{C}_\text{MIN}}=7\:\text{V}\$ to work with at the collector. The mid-point is half that, so the quiescent collector voltage should be set at \$V_{\text{C}_\text{Q}}=1\:\text{V}+2\:\text{V}+\frac{7}{2}\:\text{V}=6.5\:\text{V}\$. This gives a maximum swing of \$\pm\: 3.5\:\text{V}\$ at the collector.

From Figure 11 on the datasheet indicated above, I can see that \$V_{\text{BE}_\text{Q}}\approx 700\:\text{mV}\$ at \$I_{\text{C}_Q}=10\:\text{mA}\$. So \$V_{\text{B}_\text{Q}}\approx 1.7\:\text{V}\$.

I'll want the resistor divider current to be enough that it is stiff relative to the required base current (since the base current will vary with the signal.) This tends to mean that it is about \$\frac{1}{10}\$th of \$I_{\text{C}_Q}\$, or about \$1\:\text{mA}\$. So, \$R_\text{B1}=\frac{10\:\text{V}-1.7\:\text{V}}{1\:\text{mA}}=8.3\:\text{k}\Omega\$ and \$R_\text{B2}=\frac{1.7\:\text{V}}{1\:\text{mA}-100\:\mu\text{A}}=1.889\:\text{k}\Omega\$. Rounding these to nearby values, I decide to use \$R_\text{B1}=8.2\:\text{k}\Omega\$ and \$R_\text{B2}=1.8\:\text{k}\Omega\$.

At this point I also can compute \$R_\text{E}=\frac{1\:\text{V}}{10\:\text{mA}+100\:\mu\text{A}}\approx 99\:\Omega\$. Clearly, the nearest value is \$R_\text{E}=100\:\Omega\$. Similarly, \$R_\text{C}=\frac{10\:\text{V}-6.5\:\text{V}}{10\:\text{mA}}=350\:\Omega\$. Since I know in advance that the base current will probably be a little lower than I estimated (\$100\:\mu\text{A}\$), the base voltage will be a little higher and so the quiescent current probably just a little higher and so I will choose a slightly lower value, selecting the standard value of \$R_\text{C}=330\:\Omega\$.

This suggests a gain of about \$A_V=\frac{330\:\Omega}{100\:\Omega}\approx 3.3\$. (It will be a little less, perhaps more like \$3.2\$, because of \$r_e\approx 3\:\Omega\$.) If more gain than this is needed then an AC gain leg must be added to the emitter (a series resistor and capacitor to ground from the emitter.) However, your circuit doesn't allow for that. So that's your gain estimate.



If you are curious, you can compute the estimated base current using the following equation (derived from the Thevenin voltage law moving from the Thevenin equivalent at the base through to ground via the emitter and emitter resistor):

$$I_{\text{B}_\text{Q}}=\frac{V_\text{TH}-V_{\text{BE}_\text{Q}}}{R_\text{TH}+\left(\beta+1\right)\cdot R_\text{E}}=\frac{1.8\:\text{V}-700\:\text{mV}}{1476\:\Omega+101\cdot 100\:\Omega}\approx 95\:\mu\text{A}$$



Now it is your task to select a BJT and associated datasheet that is appropriate for the quiescent collector current you've selected. Keep in mind the power dissipation involved -- you won't be able to use a TO-92 packaged device here. But you can follow through with the above process, making your own choices here and there as you go. Then work out the resistor values.