Since the galvanometer shows no deflection, the potential difference across the resistor $R$ must be equal to the voltage of the second battery, which is $2\text{ V}$. The current through the $400\text{ }\Omega$ resistor and $R$ is the same. The voltage across the $400\text{ }\Omega$ resistor is $10\text{ V} - 2\text{ V} = 8\text{ V}$. The current $I = \frac{8\text{ V}}{400\text{ }\Omega} = 0.02\text{ A}$. Since $V_R = I \times R$, we have $2\text{ V} = 0.02\text{ A} \times R$, so $R = \frac{2}{0.02} = 100\text{ }\Omega$.