Answer
$C_{new} = 57 \mathrm{~\mu F} $
Work Step by Step
Capacitors $12 \mathrm{~\mu F}$ and $6 \mathrm{~\mu F}$ are in series, therefore their equivalent capacitor is given by
$$\dfrac{1}{{C}'} = \dfrac{1}{12\mathrm{~\mu F} } + \dfrac{1}{6 \mathrm{~\mu F} } = 0.25 $$
Take the reciprocal
$$ {C}' = 4 \mathrm{~\mu F}$$
${C}'$, $11 \mathrm{~\mu F}$ and the new capacitor $C_{new}$ are in parallel where their equivalent capacitance is calculated by
$$ {C}'' = 11 \mathrm{~\mu F} + C_{new} + {C}' = 15 \mathrm{~\mu F} +C_{new} $$
This combination $ {C}''$ is in series with $9 \mathrm{~\mu F} $ and their equivalent in series is $ {C}''' = 8 \mathrm{~\mu F} $ and we could get $C_{new}$ in the next
\begin{gather*}
\dfrac{1}{{C}''' } = \dfrac{1}{9 \mathrm{~\mu F} } + \dfrac{1}{15 \mathrm{~\mu F} +C_{new}}\\
\dfrac{1}{15 \mathrm{~\mu F} +C_{new}} = \dfrac{1}{{C}''' } - \dfrac{1}{9 \mathrm{~\mu F} } \\
\dfrac{1}{15 \mathrm{~\mu F} +C_{new}} = \dfrac{1}{72 \mathrm{~\mu F} }\\
15 \mathrm{~\mu F} +C_{new} = 72 \mathrm{~\mu F} \\
\boxed{C_{new} = 57 \mathrm{~\mu F} }
\end{gather*}
