Dielectrics question from JEE Main 2022 (28 Jul, Shift 2)

A slab of dielectric constant $K$ has the same cross-sectional area as the plates of a parallel plate capacitor and thickness $\frac{3}{4}d$, where $d$ is the separation of the plates. The capacitance of the capacitor when the slab is inserted between the plates will be:

(Given $C_0$ = capacitance of capacitor with air as medium between plates.)