$C_{1}=\frac{k_{1} \in_{0} A}{d / 3}=\frac{3 k_{1} \in_{0} A}{d}=\frac{3 \times 3 \in_{0} A}{d}=\frac{9 \in_{0} A}{d}$

and

$C_{2}=\frac{k_{2} \in_{0} A}{2 d / 3}=\frac{3 k_{2} \in_{0} A}{2 d}=\frac{3 \times 6 \in_{0} A}{2 d}=\frac{9 \in_{0} A}{d}$

The equivalent capacitance $C_{\mathrm{eq}}$ is

$\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{d}{9 \in_{0} A}+\frac{d}{9 \in_{0} A}=\frac{2 d}{9 \in_{0} A}$

$\therefore C_{e q}=\frac{9}{2} \frac{\in_{0} A}{d}=\frac{9}{2} \times 9 \,p F=40.5 \,p F$