Electrostatics: JEE Main 2022 Physics question

A slab of dielectric constant $\mathrm{K}$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4} \mathrm{~d}$, where $\mathrm{d}$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be : (Given $\mathrm{C}_{0}$ = capacitance of capacitor with air as medium between plates.)