c
\(\mathrm{U}_{\mathrm{B}}=\mathrm{U}_{\mathrm{E}}\)

\(\frac{1}{2} \mathrm{Li}^{2}=\frac{\mathrm{q}^{2}}{2 \mathrm{C}}\)

where \(q\) is the charge on the capacitor when energy stored equally between the electric and magnetic fields.

\(\therefore \mathrm{q}^{2}=\mathrm{LCi}^{2}\)          ........\((i)\)

Now maximum charge on the capacitor is \(Q\)

when maximum current  \(i_{0}\) flows through the \(L-C\) circuit.

So, maximum energy stored in the electric and magnetic fields are

\(\left(\mathrm{U}_{\mathrm{E}}\right)_{\max }=\frac{\mathrm{Q}^{2}}{2 \mathrm{C}}\) and \(\left(\mathrm{U}_{\mathrm{B}}\right)_{\max }=\frac{1}{2} \mathrm{Li}_{0}^{2}\)

Given \(\left(\mathrm{U}_{\mathrm{B}}\right)=\frac{1}{2}\left(\mathrm{U}_{\mathrm{E}}\right)_{\max }\)

\(\frac{1}{2} \mathrm{Li}^{2}=\frac{1}{2} \frac{\mathrm{Q}^{2}}{2 \mathrm{C}}=\frac{\mathrm{Q}^{2}}{4 \mathrm{C}}\)

\(\frac{{{{\rm{q}}^2}}}{{2{\rm{C}}}} = \frac{{{{\rm{Q}}^2}}}{{4{\rm{C}}}}\)      [from equation    \((i)\)]

\(\therefore \)   \(\mathrm{q}^{2}=\frac{\mathrm{Q}^{2}}{2}\)

\(\mathrm{q}=\frac{\mathrm{Q}}{\sqrt{2}}\)