A parallel plate capacitor filled with oil of dielectric constant 3...

The capacitance of a parallel plate capacitor filled with a dielectric of constant $K$ is given by $C = \dfrac{K \epsilon_0 A}{d}$.

Given $K = 3$, the initial capacitance is $C = \dfrac{3 \epsilon_0 A}{d}$.

When the oil is completely removed, the medium between the plates becomes air or vacuum, for which the dielectric constant is $1$.

The new capacitance becomes $C' = \dfrac{\epsilon_0 A}{d}$.

From the initial equation, we have $\dfrac{\epsilon_0 A}{d} = \dfrac{C}{3}$.

Therefore, the new capacitance is $C' = \dfrac{C}{3}$.

Answer: $\dfrac{\text{C}}{3}$