In figure, charge on the capacitor is plotted against potential difference across the capacitor. The capacitance and energy stored in the capacitor are respectively.

Energy stored in a capacitor is given as
\(U=\frac{1}{2} C V^{2}...(i)\)
where, \(C\) is the capacitance. From the given graph, we get

Slope \(=\tan \theta=\frac{Q}{V}=\frac{120 \mu \mathrm{C}}{10 \mathrm{~V}}...(ii)\)
As we know, capacitance is given as
\(C=\frac{Q}{V}\)
Substituting the value of \(\frac{Q}{V}\) from Eq. (ii) in the above relation, we get
\(\begin{aligned}
C &=\frac{120 \mu \mathrm{C}}{10 \mathrm{~V}}=\frac{120 \times 10^{-6}}{10}=12 \times 10^{-6} \mathrm{~F} \\
&=12 \mu \mathrm{F}
\end{aligned}\)
Now, substituting the values of \(C\) and \(V\) in Eq. (i), we get
\(\begin{aligned}
U &=\frac{1}{2} \times 12 \times(10)^{2} \\
&=600 \mu \mathrm{J}
\end{aligned}\)