In figure below, if switch $S$ is open then total energy stored in the capacitors, is suppose $E_0$
Energy in IF capacitor,
$E_{1}=\frac{1}{2} C V^{2}=\frac{1}{2} C\left(\frac{Q}{C}\right)^{2}$
$=\frac{1}{2} \frac{Q^{2}}{C}=\frac{1}{2} \times \frac{4^{2}}{1}=8 \,J$
Similarly, $E_{2}=\frac{1}{2} \frac{Q^{2}}{C}=\frac{2^{2}}{2 \times 2}=1 \,J$
So, the total energy, $E_{0}=E_{1}+E_{2}=8+ l =9 J$
Now, switch's is closed then the common potential of Capacitors,
$V_{\text {common }} =\frac{C_{1} V_{1}+C_{2} V_{2}}{C_{1}+C_{2}}$
$=\frac{1 \times 4+2 \times 1}{1+2}=\frac{6}{3}=2\, V$
Hence, now the new arrangement of energy,
$E_{1}=\frac{1}{2} C_{1} V_{\text {common }}^{2}=\frac{1}{2} \times 1 \times 4=2 \,J$
and $E_{2}=\frac{1}{2} C_{2} V_{\text {common }}^{2}=\frac{1}{2} \times 2 \times 4=4 \,J$
Now, from conservation of energy,
Energy stored in the inductor,
$E_{L} \simeq E_{0}-\left(E_{1}+E_{2}\right)$
$=9-(2+4)=3\, J$
Hence, the inductor has $3\, J$ of energy.
