$\mathrm{C}^{\prime}=\frac{\frac{\varepsilon_{\mathrm{o}} \times 3 \mathrm{A}}{\mathrm{d} / 3} \times \frac{\varepsilon_{\mathrm{o}} \times 6 \mathrm{A}}{2 \mathrm{d} / 3}}{\frac{\varepsilon_{\mathrm{o}} \times 3 \mathrm{A}}{\mathrm{d} / 3}+\frac{\varepsilon_{\mathrm{o}} \times 6 \mathrm{A}}{2 \mathrm{d} / 3}}=\frac{\frac{9 \varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}} \times \frac{9 \varepsilon_{\mathrm{o}} \mathrm{A}}{2 \mathrm{d}}}{\frac{18 \varepsilon_{\mathrm{o}} \mathrm{A}}{2 \mathrm{d}}}=\frac{9}{2} \frac{\varepsilon_{\mathrm{o}} \mathrm{A}}{\mathrm{d}}$
$C^{\prime}=\left(\frac{9}{2} \times 9\right) \,p F=40.5 \,p F$

Reason : Two equipotential surfaces are parallel to each other.

$V$. Energy stored in the above combination is $\mathrm{E}$. When these capacitors are connected in series to the same supply, the stored energy is $\frac{9}{\mathrm{x}} \mathrm{E}$. The value of $x$ is___________.
[take $\ln 5=1.6, \ln 3=1.1]$