A parallel plate air capacitor has capacity ' $C$ ' and distance of separation between plates is ' d '. If a conducting sheet of thickness $\frac{2 \mathrm{~d}}{3}$ is inserted in between the plates, the capacitance becomes $C_1$. The ratio of $\frac{C_1}{C}$ is

The original capacitance of a parallel plate capacitor is $C = \frac{\epsilon_0 A}{d}$.

When a conducting sheet of thickness $t = \frac{2d}{3}$ is inserted, the electric field only exists in the air gaps. The effective separation becomes $d' = d - t = d - \frac{2d}{3} = \frac{d}{3}$.

The new capacitance is therefore $C_1 = \frac{\epsilon_0 A}{d'} = \frac{\epsilon_0 A}{d/3} = \frac{3\epsilon_0 A}{d}$.

The ratio evaluates to $\frac{C_1}{C} = \frac{\frac{3\epsilon_0 A}{d}}{\frac{\epsilon_0 A}{d}} = 3$.

Final ratio: $\boxed{3}$