Answer
$I = 0.053~A$
Work Step by Step
In part (b), we found that $f_d = 318~Hz$
We can find the impedance:
$Z = \sqrt{R^2+X_L^2}$
$Z = \sqrt{R^2+(2\pi~f_d~L)^2}$
$Z = \sqrt{(80.0~\Omega)^2+[(2\pi)(318~Hz)(40.0\times 10^{-3}~H)]^2}$
$Z = \sqrt{(80.0~\Omega)^2+(80.0~\Omega)^2}$
$Z = 113.1~\Omega$
We can find the current amplitude:
$I = \frac{\mathscr{E}_m}{Z}$
$I = \frac{6.00~V}{113.1~\Omega}$
$I = 0.053~A$
