The space between the plates of a parallel plate capacitor of...

The initial capacitance is $C = \frac{\epsilon_0 A}{d}$.
From the figure, the capacitor is divided into three parts.
The bottom half is divided into two capacitors $C_2$ and $C_3$ in parallel, each with area $A/2$ and thickness $d/2$.
$C_2 = \frac{K_2 \epsilon_0 (A/2)}{d/2} = K_2 \frac{\epsilon_0 A}{d} = 3C$
$C_3 = \frac{K_3 \epsilon_0 (A/2)}{d/2} = K_3 \frac{\epsilon_0 A}{d} = 5C$
The equivalent capacitance of the bottom part is $C_{eq, bottom} = C_2 + C_3 = 3C + 5C = 8C$.
The top half is a single capacitor $C_1$ with area $A$ and thickness $d/2$.
$C_1 = \frac{K_1 \epsilon_0 A}{d/2} = 2K_1 \frac{\epsilon_0 A}{d} = 2(2)C = 4C$.
The total capacitance $C'$ is the series combination of $C_1$ and $C_{eq, bottom}$.
$\frac{1}{C'} = \frac{1}{C_1} + \frac{1}{C_{eq, bottom}} = \frac{1}{4C} + \frac{1}{8C} = \frac{2+1}{8C} = \frac{3}{8C}$
$C' = \frac{8}{3}C$
Comparing with $C' = \frac{n}{3}C$, we get $n = 8$.