If the galvanometer $G$ does not show any deflection in the circuit shown, the value of $R$ is given by:

As given in the question,

$I_g=0$

Potential drop across $R$ will be equal to $2 \mathrm{V}$.

Now, current through the left circuit can be written as,

$I=\frac{10}{400+R}$.

Therefore, potential drop across $R$ will be,

$$\begin{gathered} V_R=IR \\ \Rightarrow 2=\frac{10R}{400+R} \\ \Rightarrow 800+2R=10R \\ \Rightarrow R=100 \Omega \end{gathered}$$