a
Here,

Input resistance, \(R_{i}=100 \Omega\)

Change in base current, \(\Delta I_{B}=40 \mu \mathrm{A}\)

Change in collector current, \(\Delta I_{\mathrm{c}}=2 \mathrm{mA}\) 

Load resistance, \(R_{L}=4 \mathrm{k} \Omega=4 \times 10^{3} \Omega\)

Current gain, \(\beta=\frac{\Delta I_{C}}{\Delta I_{B}}=\frac{2 \mathrm{mA}}{40 \mu \mathrm{A}}=\frac{2 \times 10^{-3} \mathrm{A}}{40 \times 10^{-6} \mathrm{A}}=50\)

Voltage gain of the amplifier is

\(A_{V}=\beta \frac{R_{L}}{R_{i}}=50 \times \frac{4 \times 10^{3}}{100}=2000\)