= Solution
The ideal voltage source drives a series $L$–$R$–$C$ branch and a separate resistor $R$ in parallel. At $\omega_0=1/\sqrt{LC}$, the series branch is purely resistive, so its peak current is $I_p=V_0/R$. Its stored energies are
$$
U_L=\frac12LI_p^2\sin^2\omega_0t,\qquad U_C=\frac12LI_p^2\cos^2\omega_0t.
$$
Their sum is constant in sinusoidal steady state, with value $U_{\max}=LV_0^2/(2R^2)$. Each resistor dissipates average power $V_0^2/(2R)$, including the resistor directly across the source. Over one cycle,
$$
W_{\rm diss}=\frac{2\pi}{\omega_0}\frac{V_0^2}{R},\qquad
\boxed{\frac{U_{\max}}{W_{\rm diss}}=\frac{\omega_0L}{4\pi R}=\frac1{4\pi R}\sqrt{\frac LC}.}
$$
Omitting the parallel resistor would double this ratio. The expression concerns the driven steady state, not a switch-on transient.
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