A capacitor of capacitance 100μ F is charged to a potential of 12V...

By energy conservation, it follows that

\dfrac{1}{2}C{V}^{2} = \dfrac{1}{2}L{I}_\max^{2} . . . \left(1\right)

Equation (1) implies that

\begin{cases}{I}_\max & = & \sqrt{\dfrac{C}{L}}V \\ & = & \sqrt{\dfrac{100 \times {10}^{- 6}}{6 . 4 \times {10}^{- 3}}} \times 12 \\ & = & \dfrac{12}{8} \\ & = & 1 . 5A\end{cases}