In figure below, if switch $S$ is open then total energy stored in the capacitors, is suppose $E_0$.


Energy in IF capacitor,
$$
E_1=\frac{1}{2} C V^2=\frac{1}{2} C\left(\frac{Q}{C}\right)^2=\frac{1}{2} \frac{Q^2}{C}=\frac{1}{2} \times \frac{4^2}{1}=8 \mathrm{~J}
$$

Similarly, $E_2=\frac{1}{2} \frac{Q^2}{C}=\frac{2^2}{2 \times 2}=1 \mathrm{~J}$
So, the total energy, $E_0=E_1+E_2=8+1=9 \mathrm{~J}$ Now, switch's is closed then the common potential of Capacitors,
$$
\begin{aligned}
V_{\text {common }} & =\frac{C_1 V_1+C_2 V_2}{C_1+C_2} \\
& =\frac{1 \times 4+2 \times 1}{1+2}=\frac{6}{3}=2 \mathrm{~V}
\end{aligned}
$$
Hence, now the new arrangement of energy,
and
$$
\begin{aligned}
& E_1=\frac{1}{2} C_1 V_{\text {common }}^2=\frac{1}{2} \times 1 \times 4=2 \mathrm{~J} \\
& E_2=\frac{1}{2} C_2 V_{\text {common }}^2=\frac{1}{2} \times 2 \times 4=4 \mathrm{~J}
\end{aligned}
$$

Now, from conservation of energy,
Energy stored in the inductor, $E_L \simeq E_0-\left(E_1+E_2\right)$
$$
=9-(2+4)=3 \mathrm{~J}
$$

Hence, the inductor has $3 \mathrm{~J}$ of energy.