(A) Given: $L = 20 \; mH = 20 \times 10^{-3} \; H$,$C = 50 \; \mu F = 50 \times 10^{-6} \; F$,$Q = 10 \; mC = 10 \times 10^{-3} \; C$.
$(a)$ Total energy $E = \frac{1}{2} \frac{Q^2}{C} = \frac{(10 \times 10^{-3})^2}{2 \times 50 \times 10^{-6}} = 1 \; J$. Since resistance $R = 0$,the total energy is conserved.
$(b)$ Natural angular frequency $\omega = \frac{1}{\sqrt{LC}} = \frac{1}{\sqrt{20 \times 10^{-3} \times 50 \times 10^{-6}}} = 10^3 \; rad/s$. Frequency $f = \frac{\omega}{2\pi} = \frac{1000}{2\pi} \approx 159.2 \; Hz$.
$(c)$ Period $T = \frac{1}{f} = 2\pi \times 10^{-3} \; s \approx 6.28 \; ms$.
$(i)$ Electrical energy is max at $t = 0, \frac{T}{2}, T, \dots = n\frac{T}{2}$ for $n = 0, 1, 2, \dots$.
$(ii)$ Magnetic energy is max when electrical is zero,at $t = \frac{T}{4}, \frac{3T}{4}, \dots = (2n+1)\frac{T}{4}$ for $n = 0, 1, 2, \dots$.
$(d)$ Energy is shared equally when $U_E = U_B = \frac{E}{2}$. This occurs when $Q' = \frac{Q}{\sqrt{2}}$. Since $Q' = Q \cos(\omega t)$,we have $\cos(\omega t) = \frac{1}{\sqrt{2}}$,so $\omega t = (2n+1)\frac{\pi}{4}$. Thus $t = (2n+1)\frac{T}{8} = 0.785 \; ms, 2.355 \; ms, \dots$.
$(e)$ If a resistor is inserted,the entire initial energy of $1 \; J$ is eventually dissipated as heat due to damping.