In the given circuit, $\mathrm{C}_3$ and $\mathrm{C}_4$ are in series and
$$
\begin{aligned}
\mathrm{C}_3 & =\mathrm{C}_4=8 \mu \mathrm{F} \\
\therefore \quad \frac{1}{\mathrm{C}_5} & =\frac{1}{\mathrm{C}_3}+\frac{1}{\mathrm{C}_4}
\end{aligned}
$$
$$
\begin{aligned}
\therefore \quad \mathrm{C}_{\mathrm{s}} & =\frac{\mathrm{C}_3^2}{2 \mathrm{C}_3} \\
\mathrm{C}_{\mathrm{s}} & =\frac{\mathrm{C}_3}{2} \\
\therefore \quad \mathrm{C}_{\mathrm{S}} & =4 \mu \mathrm{F}
\end{aligned}
$$
$\mathrm{C}_5$ and $\mathrm{C}_6$ are in parallel and $\mathrm{C}_5=\mathrm{C}_6=4 \mu \mathrm{F}$
$$
\begin{aligned}
\mathrm{C}_P & =\mathrm{C}_5+\mathrm{C}_6 \\
\therefore \quad \mathrm{C}_{\mathrm{P}} & =8 \mu \mathrm{F}
\end{aligned}
$$
$\therefore \quad$ Equivalent circuit is as shown in figure.



Now, $C_2$ and $C_P$ are in series and their combination in parallel with $\mathrm{C}_5$
$$
\begin{aligned}
\therefore \quad C_E & =\frac{C_2 C_P}{C_2 C_P}+C_5 \\
& C_E=\frac{(8)(8)}{16}+4 \\
\therefore \quad C_E & =8 \mu \mathrm{F}
\end{aligned}
$$
Now, $C_1$ and $C_E$ are in series,
$$
\begin{aligned}
& \therefore \quad \mathrm{C}=\frac{\mathrm{C}_1 \mathrm{C}_{\mathrm{E}}}{\mathrm{C}_1+\mathrm{C}_{\mathrm{E}}} \\
& \therefore \quad \mathrm{C}=\frac{(8)(8)}{16} \\
& \therefore \quad \mathrm{C}=4 \mu \mathrm{F}
\end{aligned}
$$