A capacitor C1=4 μ F is charged to 4.5 V and then, at t=0, connected in a single series loop with an uncharged capacitor C2=12 μ F and an ideal…

Let $q$ be the charge that has flowed round the loop. The loop equation is $L\dfrac{dI}{dt}=V_0-\dfrac{q}{C_1}-\dfrac{q}{C_2}$, so $q$ oscillates about the equilibrium value $q_0=V_0C_{eq}$ with $C_{eq}=\dfrac{C_1C_2}{C_1+C_2}=3\ \mu\mathrm F$, and $q$ starts from $0$. The current is maximum when $dI/dt=0$, i.e. at $q=q_0$. Energy conservation: the initial energy $\tfrac12C_1V_0^2$ equals $\tfrac12LI^2+\tfrac12C_1\left(V_0-\tfrac{q_0}{C_1}\right)^2+\tfrac12\dfrac{q_0^2}{C_2}$. Expanding, the capacitor terms add up to $\tfrac12C_1V_0^2-\tfrac12V_0^2C_{eq}$, so $\tfrac12LI_{max}^2=\tfrac12V_0^2C_{eq}$. Equivalently $I_{max}=q_0\omega$ with $\omega=1/\sqrt{LC_{eq}}$. Therefore $I_{max}=V_0\sqrt{\dfrac{C_{eq}}{L}}=4.5\sqrt{\dfrac{3\times10^{-6}}{0.75}}=4.5\times2\times10^{-3}\ \mathrm A=9\ \mathrm{mA}$. (A direct numerical integration of the circuit equation confirms this.) As a sanity check, the oscillation of $q$ is simple harmonic about $q_0=13.5\ \mu\mathrm C$ with amplitude equal to $q_0$ (it starts from $q=0$ at rest), angular frequency $\omega=1/\sqrt{LC_{eq}}\approx667\ \mathrm{rad\,s^{-1}}$, so $I_{max}=q_0\omega\approx9\ \mathrm{mA}$, in agreement.