Calculation of (TC) in BJT
Your circuit has the basic structure of an NPN \$V_\text{BE}\$ multiplier. Typically, a \$V_\text{BE}\$ multiplier is operated with a current source (or sink, as in your case) and they provide a predictable voltage difference between the collector and emitter.

The basic idea is that the base-emitter junction will present a voltage difference of about \$V_\text{BE}\approx 700\:\text{mV}\$ at \$27^\circ\text{C}\$ (\$300\:\text{K}\$), which then appears across \$R_1\$. (In your case you are told a slightly different value for a slightly different temperature.) So \$V_{\text{R}_1}=V_\text{BE}\$. This implies a current in \$R_1\$ of \$I_{\text{R}_1}= \frac{V_\text{BE}}{R_1}\$, which must be supplied by a similar current in \$R_2\$.

The above paragraph I just wrote already conflicts with your writing:


  the current in R2 is 0.2mA going down and the current in R1 is 0.1mA
  going down


I'm saying instead that the current in \$R_1\$ and \$R_2\$ is about the same (only slightly different because of the BJT's base recombination current.)

The voltage drop across \$R_2\$ is \$V_{\text{R}_2}= R_2\cdot \left(I_{\text{R}_1}+I_\text{B}\right)=V_\text{BE}\cdot\frac{R_2}{R_1}+R_2\cdot I_\text{B}\$ This difference across \$R_2\$ is added to \$V_{\text{R}_1}\$ so that \$V_\text{CE}= V_\text{BE}\cdot\left(1+\frac{R_2}{R_1}\right)+R_2\cdot I_\text{B}\$.

In your circuit's case, you were able to correctly estimate the base current as \$I_\text{B}\approx 9.9\:\mu\text{A}\$ based on being provided the emitter current, \$I_\text{E}=1\:\text{mA}\$ and the fact that \$\beta=100\$. But this base current must then be added to \$I_{\text{R}_1}= \frac{V_\text{BE}}{R_1}=100\:\mu\text{A}\$ so that \$I_{\text{R}_2}=I_{\text{R}_1}+I_\text{B}=109.9\:\mu\text{A}\$.

(I honestly don't know where you got your figure for \$R_2\$'s current.)

At this point, I think you can easily compute the voltage drop across \$R_2\$ and therefore the voltage drop across both resistors. This should provide a better answer for the simpler case prior to the temperature change.

(You may also will need to work out the value of \$I\$. And I think you do know how to do that. But whether or not it is important is a question that is addressed next.)



Once you have the above worked out, correctly, and you have the value of \$I\$ there is another problem to face, now. Does \$I\$ now stay exactly the same, when the temperature changes to \$75^\circ\text{C}\$? If so, then the emitter current will change and therefore so also will the base-emitter voltage and the base recombination current. But if the emitter current is instead said to remain constant then by implication the value of \$I\$ must instead change.

The latter choice would be more simply analyzed and wouldn't require changing the assumptions you were given. But I can't tell you which to hold as true since I don't know what's expected of you.

Assuming the simpler situation (where \$I\$ changes to accommodate a fixed value for \$I_\text{E}\$), then all you need to do is work out the change in \$V_\text{BE}\$ and from there work out the current in \$R_1\$, then the current in \$R_2\$ and then the voltage drop across \$R_2\$, etc.



The \$V_\text{BE}\$ multiplier has variations, as well.

The multiplier you have lacks any adjustment for the Early Effect, for example. A resistor is often added to the collector to provide an adjustment for that purpose to handle variations in the current passing through the multiplier.

Also, the tempco can be moved around, as well: either by providing the equivalent of a voltage source in the \$R_1\$ leg of the schematic or in the \$R_2\$ leg, depending on whether you want to increase or decrease the tempco.

A variable resistor is often added to the \$R_1\$ leg to make the multiplier adjustable, too.

Just be aware that you've got a simpler case and that there are other cases with additional parts to cope with what's important to the designer at the time.