A parallel plate capacitor has plate area 100 m ^ 2 and plate...

Here, Plate area$A = 100{m}^{2}$
Using, Capacity of parallel plate capacitor is,

$C = \dfrac{k{ε}_{0}A}{d}$; where $d$ is separation between the plates and $k$ is dielectric constant.

The capacitance with dielectric will be,
${C}_{1} = \dfrac{10{ϵ}_{0} ( 100 )}{5}$
$= 200{ϵ}_{0}$

The capacitance without dielectric or air will be,
${C}_{2} = \dfrac{{ϵ}_{0} ( 100 )}{5} = 20{ϵ}_{0}$
${C}_{1} \text{ and } {C}_{2}$ are in series so,

${C}_{\text{eqv.}} = \dfrac{{C}_{1}{C}_{2}}{{C}_{1} + {C}_{2}}$

$= \dfrac{4000{ϵ}_{0}}{220}$
$= 160 . 9 \times {10}^{- 12} ≃ 161pF$