If the galvanometer G does not show any deflection in the circuit...

As given in the question,

${I}_{g} = 0$

Potential drop across $R$ will be equal to $2V$.

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

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

Therefore, potential drop across $R$ will be,

${V}_{R} = IR ⇒ 2 = \dfrac{10R}{400 + R} ⇒ 800 + 2R = 10R ⇒ R = 100Ω$