In the circuit, the galvanometer $G$ shows zero deflection. If the batteries $A$ and $B$ have negligible internal resistance, the value of the resistor $R$ will be:

Since the galvanometer shows zero deflection, the current in the $2\text{V}$ branch is zero.This implies the potential difference across the resistor $R$ is equal to the EMF of battery $B$, which is $2\text{V}$.Current in the main loop $I=\frac{12}{500+R}$.Voltage across $R$: $V_R=I\cdot R=2$.
$$\frac{12R}{500+R}=2$$
$$12R=1000+2R$$
$$10R=1000⟹R=100\Omega$$