Explanation

(c) AC power gain = $\frac{Change\ in\ output\ power}{Change\ in\ input\ power}$
= $\frac{{\mathrm{\Delta}V}_{c} \times {\mathrm{\Delta}i}_{c}}{{\mathrm{\Delta}V}_{i} \times {\mathrm{\Delta}i}_{b}} = \left( \frac{{\mathrm{\Delta}V}_{c}}{{\mathrm{\Delta}V}_{i}} \right) \times \left( \frac{{\mathrm{\Delta}i}_{c}}{{\mathrm{\Delta}i}_{b}} \right) = A_{V} \times \beta_{AC}$
where $A_{V}\ is\ voltage\ gain$ and $(\beta)_{AC}$ is AC current gain.
Also,
$A_{V} = \beta_{AC} \times resistance\ gain\ \left( = \frac{R_{o}}{R_{i}} \right)$
Given, $A_{V} = 50,\ R_{o} = 200\ \Omega,\ R_{i} = 100\ \Omega$
Hence, 50 = $\beta_{AC} \times \frac{200}{100}$
$\therefore$ $\beta_{AC} = 25$
Now, AC power gain = $A_{V} \times \beta_{AC}$ = 50 $\times$ 25 = 1250