Designing a BJT Amplifier given some constraints
First, translate the specifications into constraint equations.

For the static power dissipation:

Assume, for now, that \$I_{R2} \ge 10 \cdot I_B = \dfrac{I_C}{10}\$ for the worst case \$\beta = 100 \$.

The supply current is then:

\$I_{PS} =  I_C + 11 \cdot I_B = 1.11 \cdot I_C \$

The static power constraint then becomes:

\$\rightarrow I_C &lt; \dfrac{25mW}{1.11 \cdot 10V} = 2.25mA\$

The bias equation:

The BJT bias equation is:

\$I_C = \dfrac{V_{BB} - V_{EE} - V_{BE}}{\frac{R_{BB}}{\beta} + \frac{R_{EE}}{\alpha}} \$

For this circuit, we have:

\$V_{BB} = 10V \dfrac{R_2}{R_1 + R_2}\$

\$V_{EE} = 0V\$

\$V_{BE} = 0.6V\$

\$R_{BB} = R_1||R_2\$

\$R_{EE} = R_E\$

So, the bias equation for this circuit is:

\$I_C = \dfrac{10V \frac{R_2}{R_1 + R_2} - 0.6V}{\frac{R_1||R_2}{\beta} + \frac{R_E}{\alpha}} \$

Now, you want less than 5% variation in \$I_C\$ for \$100 \le \beta \le 800\$.  After a bit of algebra, find that this requires:

\$ \rightarrow R_E &gt; 0.165 \cdot R_1||R_2 \$

Output swing:

The positive clipping level can be shown to be:

\$v^+_O = 3V = I_C \cdot R_C||R_L \$

The negative clipping level can be shown to be about:

\$v^-_O = -3V = I_C(R_C + R_E) - 9.8V \rightarrow 6.8V = I_C(R_E + R_C)\$

Put all this together:

Choose, for example, \$I_C = 1mA \$ then:

\$R_C||10k\Omega = 3k\Omega \rightarrow R_C = 4.3k\Omega\$

\$R_E + R_C = 6.8k\Omega \rightarrow R_E = 2.5k\Omega \$

Thus, \$V_E = 2.5V\$ and \$V_B = 3.1V\$

Then,

\$R_2 = \dfrac{V_B}{10 \cdot I_B} = \dfrac{3.1V}{100\mu A} = 31k\Omega \$

\$R_1 = \dfrac{10 - V_B}{11 \cdot I_B} = \dfrac{6.9}{110\mu A} = 62.7k\Omega \$

Now, check

\$0.165 \cdot R_1||R_2 = 3.42k \Omega &gt; R_E \$

So, this doesn't meet the bias stability constraint equation we established earlier.

So run through this again (use a spreadsheet!) with larger \$I_C\$ until you've met the bias stability constraint equation.

If you can't meet the constraint with \$I_C &lt; 2.25mA \$, you may need to increase current through the base voltage divider, e.g., \$I_{R2} = 20 \cdot I_B \$ and work through the static power constraint again.



As the the correctness of the clipping level calculations above has been questioned, I simulated the circuit using values calculated from the above except that \$I_C \$ was increased to \$2mA\$ for the calculation.

The DC solution:



Driving the amplifier with a 500mV 1kHz sine wave:



Note the clipping levels are precisely +3V and -3V as designed.  The variation in \$I_C\$ is just over 5% over the range of \$\beta\$ so the next step would be to increase the multiple of base current through R2 to e.g., 20 and plug in the numbers (which does result in meeting all the constraints).