To solve this problem, we need to determine the resistance of the ammeter based on the given conditions.$\\$Given:$\\$• $0.2\%$ of the main current passes through the galvanometer.$\\$• Resistance of the galvanometer is $G. \\$Let's denote:$\\$• $I$ as the main current.$\\$• $I_g$ as the current through the galvanometer.$\\$• $S$ as the shunt resistance connected in parallel with the galvanometer.$\\$From the given information, we have:$\\ I_g = 0.002 \cdot I \\$Since the galvanometer and shunt are in parallel, the voltage across them is the same.$\\$Using Ohm's Law, the voltage across the galvanometer is:$\\ V_g = I_g \cdot G \\$The current through the shunt is:$\\ I_s = I - I_g = I - 0.002 \cdot I = 0.998 \cdot I \\$The voltage across the shunt is:$\\ V_s = I_s \cdot S \\$Since $V_g = V_s,$ we have:$\\ I_g \cdot G = I_s \cdot S \\$Substitute the values of $I_g$ and $I_s: \\ 0.002 \cdot I \cdot G = 0.998 \cdot I \cdot S \\$Cancel $I$ from both sides:$\\ 0.002 \cdot G = 0.998 \cdot S \\$Solve for $S: \\ S = \frac{0.002 \cdot G}{0.998} \\$Simplify the expression:$\\ S = \frac{2}{998} \cdot G = \frac{1}{499} \cdot G \\$The resistance of the ammeter $R_a$ is the equivalent resistance of the galvanometer and shunt in parallel:$\\ \frac{1}{R_a} = \frac{1}{G} + \frac{1}{S} \\$Substitute the value of $S: \\ \frac{1}{R_a} = \frac{1}{G} + \frac{499}{G} \\$Combine the terms:$\\ \frac{1}{R_a} = \frac{500}{G} \\$Solve for $R_a: \\ R_a = \frac{G}{500} \\$Therefore, the resistance of the ammeter is $\frac{1}{500} G. \\$This corresponds to Option 3.