The maximum current in an LC circuit can be found using the following formula related to simple harmonic motion:

$I_{\text{max}} = \omega Q_{\text{max}}$,

where:

• $I_{\text{max}}$ is the maximum current,
• $\omega$ is the angular frequency, and
• $Q_{\text{max}}$ is the maximum charge on the capacitor.

The angular frequency $\omega$ for an LC circuit is given by:

$\omega = \frac{1}{\sqrt{LC}}$,

where:

• $L$ is the inductance, and
• $C$ is the capacitance.

Given that $L = 75 \, \text{mH} = 75 \times 10^{-3} \, \text{H}$, $C = 1.2 \, \mu \text{F} = 1.2 \times 10^{-6} \, \text{F}$,

and $Q_{\text{max}} = 2.7 \, \mu \text{C} = 2.7 \times 10^{-6} \, \text{C}$,

we can substitute these values into the formulas to find $I_{\text{max}}$:

$\omega = \frac{1}{\sqrt{(75 \times 10^{-3})(1.2 \times 10^{-6})}} = 3333.33 \, \text{rad/s}$,

$I_{\text{max}} = \omega Q_{\text{max}} = 3333.33 \times 2.7 \times 10^{-6} = 0.009 \, \text{A}$.

Therefore, the maximum current in the circuit is $0.009 \, \text{A}$, or equivalently, $9 \, \text{mA}$.