Answer
$C_3 = 1.67 \mathrm{~\mu F}$
Work Step by Step
From the value of the stored energy, we could get the equivalent capacitance for all the network by
\begin{gather*}
U = \dfrac{1}{2} C_{eq}V^2\\
C_{eq} = \dfrac{2U}{V^2} \\
C_{eq} = \dfrac{2\times 2.90 \times 10^{-3} \,\text{J}}{(48 \,\text{V})^2} \\
C_{eq} = 2.52 \mathrm{~\mu F}
\end{gather*}
$C_4$ is in series with the combination of $C_{123}$, so we could get the next
\begin{gather*}
\dfrac{1}{C_{eq}} = \dfrac{1}{C_{123}} + \dfrac{1}{C_4}\\
\dfrac{1}{C_{123}} = \dfrac{1}{C_{eq}} - \dfrac{1}{C_4} \\
\dfrac{1}{C_{123}} = \dfrac{1}{2.52 \mathrm{~\mu F}} - \dfrac{1}{8\mathrm{~\mu F}}\\
\dfrac{1}{C_{123}} = 0.271 \\
C_{123} = 3.67 \mathrm{~\mu F}
\end{gather*}
$C_1$ and $C_2$ are in series, so thier comination is given by
\begin{gather*}
\dfrac{1}{C_{12}} = \dfrac{1}{C_{1}} + \dfrac{1}{C_2}\\
\dfrac{1}{C_{12}} = \dfrac{1}{4\mathrm{~\mu F}} + \dfrac{1}{4 \mathrm{~\mu F}}\\
\dfrac{1}{C_{12}} = 0.5 \\
C_{12} = 2\mathrm{~\mu F}
\end{gather*}
$C_3$ is in parallel with $C_{12}$, so it is given by
\begin{gather*}
C_{123} = C_{12} + C_3 \\
C_3 = C_{123} - C_{12}\\
C_3 = 3.67\mathrm{~\mu F} - 2 \mathrm{~\mu F}\\
\boxed{C_3 = 1.67 \mathrm{~\mu F}}
\end{gather*}
