Answer
$T = 4.61\times 10^{-5}~s$
Work Step by Step
We can find the maximum charge $Q$ on the capacitor:
$Q = V~C = (0.250~V)(220\times 10^{-9}~F) = 55.0~nC$
We can find $\omega$:
$\omega = \frac{I}{Q} = \frac{7.50\times 10^{-3}~A}{55.0\times 10^{-9}~C} = 1.364\times 10^5~rad/s$
We can find the period of oscillation:
$T = \frac{2\pi}{\omega} = \frac{2\pi}{1.364\times 10^5~rad/s} = 4.61\times 10^{-5}~s$
