Given data

• Inductance: $L = 75\,\text{mH} = 75 \times 10^{-3}\,\text{H}$
• Capacitance: $C = 1.2\,\mu\text{F} = 1.2 \times 10^{-6}\,\text{F}$
• Maximum charge: $Q_{\max} = 2.7\,\mu\text{C} = 2.7 \times 10^{-6}\,\text{C}$

Use energy conservation in an LC circuit

In an ideal LC circuit, the total energy remains constant.

At maximum charge, all energy is stored in the capacitor: $U_C = \frac{Q_{\max}^2}{2C}$

At maximum current, all energy is stored in the inductor: $U_L = \frac{1}{2}LI_{\max}^2$

Equating both, $\frac{Q_{\max}^2}{2C} = \frac{1}{2}LI_{\max}^2$

Solve for maximum current

$I_{\max} = \frac{Q_{\max}}{\sqrt{LC}}$

Substitute values

First compute $LC$: $LC = (75 \times 10^{-3})(1.2 \times 10^{-6})$ $LC = 90 \times 10^{-9} = 9 \times 10^{-8}$

Then, $\sqrt{LC} = \sqrt{9 \times 10^{-8}} = 3 \times 10^{-4}$

Now, $I_{\max} = \frac{2.7 \times 10^{-6}}{3 \times 10^{-4}}$ $I_{\max} = 0.9 \times 10^{-2} = 9 \times 10^{-3}\,\text{A}$

Convert to mA

$9 \times 10^{-3}\,\text{A} = 9\,\text{mA}$

Final answer

$\boxed{9}$

Comparison with stored answer

Stored correct answer = $9$

My derived answer also = $9$, so they agree.