#### Answer

The least possible value of $t$ is $0.04$ seconds.

#### Work Step by Step

$$i=I_{max}\sin(2\pi ft)$$
For $i=I_{max},f=60$, we have
$$i=i\sin(2\pi\times60t)$$ (since $i=I_{max}$, we can replace $I_{max}$ with $i$)
$$i=i\sin(120\pi t)$$
Because $I_{max}$ is the maximum current, $I_{max}\ne0$. And as $i=I_{max}$, $i\ne0$.
So now we can divide both sides of the equation by $i$:
$$\sin(120\pi t)=1$$
$t$ refers to time, so as a rule, $t\in[0,\infty)$
Also $t$ is minimum when $120\pi t$ is minimum, as $120\pi$ is a constant.
Therefore, we would be able to find the least of $t$ as we find the first value of $120\pi t$ that would have $\sin(120\pi t)=1$ over the interval $[0,2\pi)$.
Such a value of $120\pi t$ can be found using the inverse function for sine:
$$120\pi t=\sin^{-1}1$$
$$120\pi t=\frac{\pi}{2}$$
$$t=\frac{\pi}{2\times120\pi}$$
$$t=\frac{1}{240}\approx0.04(seconds)$$
Therefore, the least possible value of $t$ satisfying the given data is $0.04$ seconds.