Answer
${\bf 1.0}\;\rm \mu F$
Work Step by Step
Recalling that the dynamics of the charge on the capacitor in $LC$-circuits is given by
$$Q=Q_0\cos(\omega t)$$
so the current in this circuit is given by
$$I=-\dfrac{ dQ}{dt}=-\dfrac{d}{dt}Q_0\cos(\omega t)$$
$$I= Q_0\omega \sin(\omega t)$$
where $Q_0 \omega =I_0$ where $I_0$ is the maximum current.
Hence,
$$I_0= Q_0 \omega $$
when $Q_0=(\Delta V_C)_{\rm max} C$, and $\omega=\sqrt{1/LC}$,
$$I_0=\dfrac{ (\Delta V_C)_{\rm max} C}{\sqrt{LC}} $$
$$I_0=\dfrac{ (\Delta V_C)_{\rm max} \sqrt{C}}{\sqrt{L }} $$
Solving for $C$,
$$C=\left[\dfrac{I_0\sqrt{L}}{(\Delta V_C)_{\rm max}}\right]^2$$
Plug the known;
$$C=\left[\dfrac{(0.60)\sqrt{(10\times 10^{-3})}}{(60)}\right]^2$$
$$C=\color{red}{\bf 1.0}\;\rm \mu F$$
