Since the exact circuit diagram is missing from the question text, we use the standard values associated with this specific NEET 2018 question: base resistance $R_B = 500 \text{ k}\Omega$, collector resistance $R_C = 4 \text{ k}\Omega$, and supply voltage $V_{CC} = 20 \text{ V}$.

1. For the input (base) loop: Applying Kirchhoff's voltage law: $V_i - I_B R_B - V_{BE} = 0$ $I_B = \frac{V_i - V_{BE}}{R_B} = \frac{20 - 0}{500 \times 10^3 \text{ } \Omega}$ $I_B = 40 \times 10^{-6} \text{ A} = 40 \text{ } \mu\text{A}$

2. For the output (collector) loop: Applying Kirchhoff's voltage law: $V_{CC} - I_C R_C - V_{CE} = 0$ Assuming $V_{CC} = V_i = 20 \text{ V}$ (as per standard configurations for this problem): $I_C = \frac{V_{CC} - V_{CE}}{R_C} = \frac{20 - 0}{4 \times 10^3 \text{ } \Omega}$ $I_C = 5 \times 10^{-3} \text{ A} = 5 \text{ mA}$

3. Current Gain ($\beta$): The common-emitter current gain $\beta$ is the ratio of collector current to base current: $\beta = \frac{I_C}{I_B} = \frac{5 \times 10^{-3} \text{ A}}{40 \times 10^{-6} \text{ A}}$ $\beta = \frac{5000}{40} = 125$