To solve this problem, we will use the concept of a balanced Wheatstone bridge.$\\$Given:$\\$• A 10 V battery in series with a 400 $\Omega$ resistor.$\\$• A 2 V battery in parallel with resistor $R. \\$• Galvanometer G shows no deflection, indicating no current through it.$\\$When the galvanometer shows no deflection, the potential difference across it is zero.$\\$This means the potential at both ends of the galvanometer is the same.$\\$Let's denote:$\\$• $V_A$ as the potential at the junction of the 10 V battery and the 400 $\Omega$ resistor.$\\$• $V_B$ as the potential at the junction of the 2 V battery and resistor $R. \\$Since the galvanometer shows no deflection:$\\ V_A = V_B \\$Calculate $V_A: \\ V_A = 10 - I \cdot 400 \\$where $I$ is the current through the 400 $\Omega$ resistor.$\\$Calculate $V_B: \\ V_B = 2 \\$Since $V_A = V_B: \\ 10 - I \cdot 400 = 2 \\$Solve for $I: \\ I \cdot 400 = 10 - 2 \\ I \cdot 400 = 8 \\ I = \frac{8}{400} \\ I = \frac{1}{50}$ A$\\$Now, calculate the voltage drop across $R: \\ V_R = I \cdot R \\ V_R = \frac{1}{50} \cdot R \\$Since the potential difference across the galvanometer is zero:$\\ V_R = 2 \\$Substitute the known values:$\\ \frac{1}{50} \cdot R = 2 \\$Solve for $R: \\ R = 2 \cdot 50 \\ R = 100 \Omega \\$Therefore, the value of $R$ is $100 \Omega$. This corresponds to Option 3.