Explanation

(b) Initially, the capacitance of capacitor
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$\ \ \ \ \ \ \ \ \ \ \ \ \ C = \frac{\varepsilon_{0}A}{d}$
$\therefore\ \ \ \ \ \ \frac{\varepsilon_{0}A}{d} = 1\mu F\ \ $ …(i)
When it is filled with dielectric of dielectric constant $K_{1}\ and\ K_{2}$ as shown, then there are two capacitors connected is parallel. So,
$\ \ \ \ \ \ \ \ \ \ \ \ \ C^{''} = \frac{K_{1}\varepsilon_{0}\left( \frac{A}{2} \right)}{d} + \frac{K_{2}\varepsilon_{0}\left( \frac{A}{2} \right)}{d}$
(as area becomes half)
$\ \ \ \ \ \ \ \ \ \ \ \ \ C''\ = \frac{4\varepsilon_{0}A}{2d} + \frac{6\varepsilon_{0}A}{2d} = \frac{2\varepsilon_{0}A}{d} + \frac{3\varepsilon_{0}A}{d}$
Using Eq. (i), we obtain
$\ \ \ \ \ \ \ \ \ \ \ \ \ C^{''} = \ 2 \times 1 + 3 \times 1 = 5\ \mu F$