A parallel plate capacitor having cross-sectional area A and...

First, recall the formula of capacity of a parallel plate capacitor at free space in terms of cross-section area and separation between plates,${C}_{0} = \dfrac{{ϵ}_{0}A}{d}$, when~ a dielectric slabs of thickness,& $t$~and dielectric constant& $k$, inserted between plates then-new capacitance becomes,~

${C}_{k} = \dfrac{{ϵ}_{0}A}{d - t + \dfrac{t}{k}}⇒ {C}_{k} = \dfrac{{ϵ}_{0}A}{d - \dfrac{d}{2} + \dfrac{d}{8}}$, here~$t = \dfrac{d}{2}$.

$⇒ {C}_{k} = \dfrac{8}{5}\dfrac{{ϵ}_{0}A}{d}⇒ {C}_{k} = \dfrac{8}{5}{C}_{0}⇒ \dfrac{Ck}{C0} = \dfrac{8}{5}$