A capacitor of 12 µF is charged to 30 V…

Step 1. The charge. $q_0 = CV_0 = 12 \times 10^{-6}\ \text{F} \times 30\ \text{V} = 3.6 \times 10^{-4}\ \text{C} = 360\ \mu\text{C}$.

Step 2. The energy. $U = \tfrac{1}{2}CV_0^2 = \tfrac{1}{2} \times 12 \times 10^{-6}\ \text{F} \times (30\ \text{V})^2 = 5.4 \times 10^{-3}\ \text{J} = 5.4$ mJ.

Step 3. The angular frequency. $LC = 0.30\ \text{H} \times 12 \times 10^{-6}\ \text{F} = 3.6 \times 10^{-6}\ \text{s}^2$, so $\omega_0 = \dfrac{1}{\sqrt{3.6 \times 10^{-6}\ \text{s}^2}} = 527$ rad/s.

Step 4. The largest current. $I_0 = \omega_0q_0 = 527\ \text{rad/s} \times 3.6 \times 10^{-4}\ \text{C} = 0.190$ A.