Explanation

(d) $C = \frac{\varepsilon_{0}A}{d}$ and $C^{'} = \frac{\varepsilon_{0}A}{2d} + \frac{\varepsilon_{0}(5A)}{2d}$
$= \frac{\varepsilon_{0}A}{2d}(1 + 5) = \frac{6\varepsilon_{0}A}{2d} = \frac{3\varepsilon_{0}A}{d}$
$\Rightarrow \mathrm{\Delta}C = C^{'} - C = \frac{3\varepsilon_{0}A}{d} - \frac{\varepsilon_{0}A}{d}$
$= \frac{2\varepsilon_{0}A}{d}$
Percentage change in capacitance
$\frac{\mathrm{\Delta}C}{C} = \frac{\frac{2\varepsilon_{0}A}{d}}{\frac{\varepsilon_{0}A}{d}} \times 100\% = 200\%$