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

If the galvanometer shows zero deflection, no current flows through the branch containing the $2\text{V}$ battery.This means the potential difference across the resistor $R$ must be exactly equal to the EMF of the smaller battery ($2\text{V}$).The total current $I$ from the $10\text{V}$ battery flows through the $400\Omega$ and $R$ resistors in series.
$$I=\frac{10}{400+R}$$
Potential across $R$:
$$V_R=I\times R=2\text{V}$$
$$\frac{10R}{400+R}=2$$
$$10R=800+2R$$
$$8R=800⟹R=100\Omega$$