\(\therefore \quad\) Collector current, \(I_{C}=\frac{(20-0)}{4 \times 10^{3}}\)

or \(I_{C}=5 \times 10^{-3} \mathrm{A}=5\, \mathrm{mA}\)

Input voltage, \(V_{i}=V_{B E}+I_{B} R_{B}\)

or \(\quad V_{i}=0+I_{B} R_{B}\)     or       \(20=I_{B} \times 500 \times 10^{3}\)

\(\therefore \quad I_{B}=\frac{20}{500 \times 10^{3}}=40\, \mu \mathrm{A}\)

\(\therefore \)        Current gain, \(\beta=\frac{I_{C}}{I_{B}}=\frac{5 \times 10^{-3}}{40 \times 10^{-6}}=125\)