d
The given capacitance is equal to two capacitances connected in series where

\(C_{1}=\frac{k_{1} \epsilon_{0} A}{d / 3}=\frac{3 k_{1} \epsilon_{0} A}{d}=\frac{3 \times 3 \epsilon_{0} A}{d}=\frac{9 \epsilon_{0} A}{d}\)

and

\(C_{2}=\frac{k_{2} \epsilon_{0} A}{2 d / 3}=\frac{3 k_{2} \epsilon_{0} A}{2 d}=\frac{3 \times 6 \epsilon_{0} A}{2 d}=\frac{9 \epsilon_{0} A}{d}\)

The equivalent capacitance \(C_{\mathrm{eq}}\) is

\(\frac{1}{C_{\mathrm{eq}}}=\frac{1}{C_{1}}+\frac{1}{C_{2}}=\frac{d}{9 \epsilon_{0} A}+\frac{d}{9 \epsilon_{0} A}=\frac{2 d}{9 \epsilon_{0} A}\)

\(\therefore C_{e q}=\frac{9}{2} \frac{\epsilon_{0} A}{d}=\frac{9}{2} \times 9 \,p F=40.5 \,p F\)