Answer
The current leads the emf.
Work Step by Step
We can write an expression for the current:
$i = I~sin~(\omega_d t-\phi)$
In this case, we can see that $\phi = -42.0^{\circ}$
When $\phi \lt 0$, the current phasor rotates ahead of the phasor $\mathscr{E}_m$
That is, the current leads the emf.
