To solve this problem, we need to find the voltage across the secondary coil and the current in the primary coil of a transformer.$\\$Given:$\\$• Efficiency of the transformer $\eta = 90\% = 0.9 \\$• Primary voltage $V_p = 200$ V$\\$• Power supply $P = 3$ kW$= 3000$ W$\\$• Current in the secondary coil $I_s = 6$ A$\\$We know that the efficiency of a transformer is given by:$\\ \eta = \frac{P_s}{P_p} \\$where $P_s$ is the power in the secondary coil and $P_p$ is the power in the primary coil.$\\$Since $P_p = 3000$ W, we can find $P_s: \\ P_s = \eta \cdot P_p = 0.9 \cdot 3000 = 2700$ W$\\$The power in the secondary coil is also given by:$\\ P_s = V_s \cdot I_s \\$Substitute the known values to find $V_s: \\ 2700 = V_s \cdot 6 \\ V_s = \frac{2700}{6} = 450$ V$\\$Now, let's find the current in the primary coil $I_p. \\$The power in the primary coil is given by:$\\ P_p = V_p \cdot I_p \\$Substitute the known values to find $I_p: \\ 3000 = 200 \cdot I_p \\ I_p = \frac{3000}{200} = 15$ A$\\$Therefore, the voltage across the secondary coil is $450$ V and the current in the primary coil is $15$ A.$\\$This corresponds to Option 2.