Answer
$I=1.72$ mA
Work Step by Step
We know that,
$\frac{1}{2}LI^2=\frac{1}{2}\frac{Q^2}{C}$
This can be rearranged as:
$LI^2=\frac{Q^2}{C}$
$I=Q\sqrt\frac{1}{LC}$
We plug in the known values to obtain:
$I=3\times 10^{-9}\sqrt\frac{1}{3\times 10^{-3}(1\times 10^{-9})}=1.7\times 10^{-3}=1.72$ mA
