In a parallel plate capacitor set up, the plate area of capacitor...

The given system can be considered as a series combination of two capacitors. The equivalent capacitance can be given as,

$\dfrac{1}{C} = \dfrac{1}{{C}_{1}} + \dfrac{1}{{C}_{2}} ⇒ \dfrac{1}{C} = \dfrac{\dfrac{d}{2}}{K{ε}_{0}A} + \dfrac{\dfrac{d}{2}}{{ε}_{0}A}$

$⇒ C = \dfrac{{ε}_{0}A}{\dfrac{d}{2K} + \dfrac{d}{2}} = \dfrac{2{ε}_{0}A}{\dfrac{d}{K} + d}$

$= \dfrac{2 \times 2{ε}_{0}}{\dfrac{1}{3 . 2} + 1} = \dfrac{4 \times 3 . 2}{4 . 2}{ε}_{0}$

$= 3 . 04{ε}_{0}$