Explanation

(b) The capacitance $C$ of a capacitor of area $A$ and distance between plates is $d$, then
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$C = \frac{\varepsilon_{0}A}{d}$
When a dielectric slab of thickness $t$ is placed between the plates, we have
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{''} = \frac{\varepsilon_{0}A}{d - t + \frac{t}{K}}$
Given, $C = 20\mu F = 20 \times 10^{- 6}F,$
$\ \ \ \ \ \ \ \ \ \ \ \ \ d = 2\ mm = 2 \times 10^{- 3}m,\ t = 1mm$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1 \times 10^{- 3}m,\ K = 2$
$\therefore\ \ \ \ \ \ \ \ \frac{C^{''}}{C} = \frac{d}{d - t\left( 1 - \frac{1}{K} \right)}\ $
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{2 \times 10^{- 3}}{2 \times 10^{- 3} - 1 \times 10^{- 3}\left( 1 - \frac{1}{2} \right)} = 1.33$
$\Rightarrow \ \ \ \ \ C^{''} = \ 1.33 \times 20 \times 10^{- 6}\ \ = 26.6\ \mu F$