When it is half filled by a dielectric of dielectric constant $K$, then

$C_{1}=\frac{K \varepsilon_{0} A}{d / 2}=2 K \frac{\varepsilon_{0} A}{d}$

and $C_{2}=\frac{\varepsilon_{0} A}{d / 2}=\frac{2 \varepsilon_{0} A}{d}$

$\therefore \frac{1}{{{C^\prime }}} = \frac{1}{{{C_1}}} + \frac{1}{{{C_2}}} = \frac{d}{{2{\varepsilon _0}A}}\left( {\frac{1}{K} + 1} \right)$ 

$ = \frac{d}{{2{\varepsilon _0}A}}\left( {\frac{1}{5} + 1} \right) = \frac{6}{{10}}\frac{d}{{{\varepsilon _0}A}}$

$C = \frac{{5{\varepsilon _0}A}}{{3d}}$

Hence, $\%$ increase in capacitance

${ = \left( {\frac{{\frac{5}{3}\frac{{{\varepsilon _0}A}}{d} - \frac{{{\varepsilon _0}A}}{d}}}{{\frac{{{\varepsilon _0}A}}{d}}}} \right) \times 100}$

${ = \left( {\frac{5}{3} - 1} \right) \times 100 = \frac{2}{3} \times 100 = 66.6\% }$