Answer
a.) $\Delta V_L = -I_0L\omega\cos(\omega t)$
b.) $I_0 = -1.27\cdot10^{-3}$ A
Work Step by Step
Assuming that "the potential difference across an inductor is measured along the direction of the current" flow,
a.) $\displaystyle \Delta V_L = -L\frac{dI}{dt}\qquad\qquad$ given that $\qquad I(t) = I_0\sin(\omega t)$
$\Delta V_L = -L[I_0\cos(\omega t)(\omega)]$
$\Delta V_L = -I_0L\omega\cos(\omega t)$
b.)
The angular frequency $\omega$ can be found by $\omega = 2\pi f$
Also, when we have the maximum voltage, we know that $\cos(\omega t)$ must be a maximum and its maximum value is always 1. $\qquad\qquad\rightarrow \cos_{max}(\omega t) = 1$
Isolating for $I_0$ we get
$I_0 = \displaystyle -\frac{\Delta V_L}{L\omega\cos(\omega t)} = -\frac{\Delta V_{L, max}}{L(2\pi f)(1)}$
$I_0 = \displaystyle -\frac{20}{(50\cdot10^{-6})(2\pi\cdot500\cdot10^3)} = -1.27\cdot10^{-3}$ A
