Answer
(a) $L = 2.36\times 10^{-5}~H$
(b) $I_L = 165~\mu A$
Work Step by Step
(a) We can find the inductance:
$I_L = \frac{V_L}{\omega~L}$
$L = \frac{V_L}{\omega~I_L}$
$L = \frac{V_L}{2\pi ~f~I_L}$
$L = \frac{2.2~V}{(2\pi) (45\times 10^6~Hz)~(330\times 10^{-6}~A)}$
$L = 2.36\times 10^{-5}~H$
(b) We can find the peak current:
$I_L = \frac{V_L}{\omega~L}$
$I_L = \frac{V_L}{2\pi ~f~L}$
$I_L = \frac{2.2~V}{(2\pi) (90\times 10^6~Hz)~(2.36\times 10^{-5}~H)}$
$I_L = 1.65\times 10^{-4}~A$
$I_L = 165~\mu A$