As shown in the figure, two parallel plate capacitors having equal...

The given data is

$A = 200 \times {10}^{- 4}{m}^{2}d = 5 \times {10}^{- 3}m$

The value of $a + b = 4 \times {10}^{- 3}m$

The formula for a parallel plate capacitor is

$C = \dfrac{{ε}_{0}A}{D}$

Let the capacitor with gap $a$ be ${C}_{1}$ and capacitor with gap $b$ be ${C}_{2}$

${C}_{1} = \dfrac{{ε}_{0}A}{a}, {C}_{2} = \dfrac{{ε}_{0}A}{b}$

The equivalent capacitance is

$\dfrac{1}{{C}_{eq}} = \dfrac{a}{{ε}_{0}A} + \dfrac{b}{{ε}_{0}A} ⇒ {C}_{eq} = \dfrac{{ε}_{0}A}{a + b}$

It is given that ${C}_{eq} = x{ε}_{0}F$

So,

$\dfrac{{ε}_{0}A}{a + b} = x{ε}_{0} ⇒ x = \dfrac{A}{a + b} ⇒ x = \dfrac{200 \times {10}^{- 4}}{4 \times {10}^{- 3}} = 5$