(N/A) From the graph in figure $(2)$, for the operating point $Q$, we have $I_B = 30 \, \mu A$, $I_C = 4 \, mA = 4 \times 10^{-3} \, A$, and $V_{CE} = 8 \, V$.
For the output circuit:
$V_{CC} = I_C R_C + V_{CE}$
$16 = (4 \times 10^{-3}) R_C + 8$
$R_C = \frac{8}{4 \times 10^{-3}} = 2 \times 10^3 \, \Omega = 2 \, k\Omega$.
For the input circuit:
$V_{BB} = I_B R_B + V_{BE}$
$16 = (30 \times 10^{-6}) R_B + 0.7$
$R_B = \frac{15.3}{30 \times 10^{-6}} = 0.51 \times 10^6 \, \Omega = 510 \, k\Omega$.
Voltage gain $A_V = \beta \frac{R_C}{R_{in}}$. Assuming input resistance $R_{in} \approx \frac{V_{BE}}{I_B} = \frac{0.7}{30 \times 10^{-6}} \approx 23.33 \, k\Omega$.
$A_V = \left( \frac{I_C}{I_B} \right) \left( \frac{R_C}{R_{in}} \right) = \left( \frac{4 \times 10^{-3}}{30 \times 10^{-6}} \right) \left( \frac{2000}{23333} \right) \approx 133.33 \times 0.0857 \approx 11.43$.
Power gain $A_P = A_V \times \beta = 11.43 \times 133.33 \approx 1524$.