A series circuit has L = 40 mH, C = 0.25 µF and R = 25 Ω…

Step 1. The resonant frequency. $LC = 40 \times 10^{-3}\ \text{H} \times 0.25 \times 10^{-6}\ \text{F} = 1.0 \times 10^{-8}\ \text{s}^2$. So $\omega_0 = \dfrac{1}{\sqrt{LC}} = \dfrac{1}{1.0 \times 10^{-4}\ \text{s}} = 10\,000\ \text{rad/s}$. Then $f_0 = \dfrac{\omega_0}{2\pi} = 1592$ Hz.

Step 2. The current. At resonance $Z = R$, so $I = \dfrac{V}{R} = \dfrac{12\ \text{V}}{25\ \Omega} = 0.48$ A.

Step 3. The coil voltage. $X_L = \omega_0L = 10\,000\ \text{rad/s} \times 40 \times 10^{-3}\ \text{H} = 400\ \Omega$. So $V_L = IX_L = 0.48\ \text{A} \times 400\ \Omega = 192$ V.

Step 4. The quality factor. $Q = \dfrac{V_L}{V} = \dfrac{192\ \text{V}}{12\ \text{V}} = 16$. Also $Q = \dfrac{\omega_0L}{R} = \dfrac{400\ \Omega}{25\ \Omega} = 16$.