To solve this problem, we need to find the voltage gain and power gain of the amplifier.$\\$Given:$\\$• Load resistance $R_L = 800 \, \Omega \\$• Voltage drop across $R_L = 0.8 \,$V$\\$• Current amplification factor ($\beta) = 0.96 \\$• Input resistance $R_{in} = 192 \, \Omega \\$First, calculate the collector current $I_C \newline I_C = \frac{Voltage \ drop \ across R_L}{R_L} = \frac{0.8}{800} = 0.001 \,$A$\, (1 \,$mA$) \\$The base current $I_B$ is given by:$\\ I_B = \frac{I_C}{\beta} = \frac{0.001}{0.96} \approx 0.001042 \,$A$\\$The input voltage $V_{in}$ is:$\\ V_{in} = I_B \cdot R_{in} = 0.001042 \cdot 192 \approx 0.2 \,$V$\\$The voltage gain $A_v$ is calculated as:$\newline A_v = \frac{Voltage \ drop \ across \ R_L}{V_{in}} = \frac{0.8}{0.2} = 4 \\$Next, calculate the power gain $A_p: \\ A_p = \frac{Output \ power}{Input \ power} \\$Output power $P_{out} = I_C^2 \cdot R_L = (0.001)^2 \cdot 800 = 0.0008 \,$W$\\$Input power $P_{in} = I_B^2 \cdot R_{in} = (0.001042)^2 \cdot 192 \approx 0.000208 \,$W$\\ A_p = \frac{P_{out}}{P_{in}} = \frac{0.0008}{0.000208} \approx 3.84 \\$Therefore, the voltage gain is 4 and the power gain is 3.84.$\\$This corresponds to Option 3.