Explanation

(b) Initial capacitance
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C = \frac{\varepsilon_{0}A}{d}$
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When it is half filled by a dielectric of dielectric constant $K$, then
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C_{1} = \frac{K\varepsilon_{0}A}{d/2} = 2K\frac{\varepsilon_{0}A}{d}$
and $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C_{2} = \frac{\varepsilon_{0}A}{d/2} = \frac{2\varepsilon_{0}A}{d}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \frac{1}{C''} = \frac{1}{C_{1}} + \frac{1}{C_{2}}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{d}{2\varepsilon_{0}A}\left( \frac{1}{K} + 1 \right) = \frac{d}{2\varepsilon_{0}A}\left( \frac{1}{5} + 1 \right)$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{6}{10}\ \frac{d}{\varepsilon_{0}A}$
$\therefore\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ C^{''} = \frac{5\varepsilon_{0}A}{3d}$
Hence, increase in capacitance
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \ \frac{\frac{5}{3}\frac{\varepsilon_{0}A}{d} - \frac{\varepsilon_{0}A}{d}}{\frac{\varepsilon_{0}A}{d}} = \frac{5}{3} - 1 = \frac{2}{3} = 66.6\%\ $