The voltage amplification \( A_v \) for a CE transistor amplifier is given by:
\[
A_v = \frac{V_{\text{out}}}{V_{\text{in}}} = \beta \cdot \frac{R_{\text{out}}}{R_{\text{in}}}
\]
Given:
- \( V_{\text{out}} = 5 \, \text{V} \),
- \( R_{\text{out}} = 4 \, \text{k}\Omega \),
- \( R_{\text{in}} = 2 \, \text{k}\Omega \),
- \( \beta = 100 \).
Using the formula:
\[
A_v = \frac{5}{V_{\text{in}}} = 100 \cdot \frac{4 \, \text{k}\Omega}{2 \, \text{k}\Omega} = 200
\]
Now, solving for \( V_{\text{in}} \):
\[
V_{\text{in}} = \frac{5}{200} = 0.025 \, \text{V} = 25 \, \text{mV}
\]
Thus, the correct answer is (B).