A series circuit has L = 0.10 H and R = 10 Ω…

Step 1. The capacitance. We want $C$, so $C = \dfrac{1}{\omega_0^2L} = \dfrac{1}{(5000\ \text{rad/s})^2 \times 0.10\ \text{H}} = 4.0 \times 10^{-7}\ \text{F} = 0.40\ \mu\text{F}$.

Step 2. The quality factor. $Q = \dfrac{\omega_0L}{R} = \dfrac{5000\ \text{rad/s} \times 0.10\ \text{H}}{10\ \Omega} = 50$.

Step 3. The bandwidth. $\Delta\omega = \dfrac{R}{L} = \dfrac{10\ \Omega}{0.10\ \text{H}} = 100\ \text{rad/s}$. Check: $\dfrac{\omega_0}{Q} = \dfrac{5000}{50} = 100$ rad/s.

Step 4. The capacitor voltage. $V_C = QV = 50 \times 2.0\ \text{V} = 100$ V.