Answer
$t = 11.2~ms$
Work Step by Step
$i(t) = I~sin~(\omega_d t-\frac{3\pi}{4})$
The current reaches a maximum when $~~sin~(\omega_d t-\frac{3\pi}{4}) = 1$
We can find $t$:
$sin~(\omega_d t-\frac{3\pi}{4}) = 1$
$\omega_d t-\frac{3\pi}{4} = \frac{\pi}{2}$
$\omega_d t = \frac{5\pi}{4}$
$t = \frac{5\pi}{4~\omega_d}$
$t = \frac{5\pi}{(4)~(350~rad/s)}$
$t = 11.2~ms$
