Answer:
Option C
Explanation:
Given, $\triangle I_{C}= 1mA=10^{-3}$ A
$\triangle I_{b}= 15\mu A=15\times 10^{-6} A$
$R_{L}= 5k$ Ω =5 x 103 Ω
Ri= 330 Ω
the voltage gain of an amplifier
$A_{r}=\frac{\triangle I_{c}\times R_{L}}{\triangle I_{b}\times R_{i} }$
= $\frac{10^{-3}\times5\times 10^{3}}{15 \times 10^{-6}\times330}=1010$