Answer
The amplitude of the voltage across the inductor is $~~1000~V$
Work Step by Step
At resonance, $\omega_d = \frac{1}{\sqrt{L~C}}$ and $I = \frac{\mathscr{E}_m}{R}$
We can find the amplitude of the voltage across the inductor:
$V_L = I_L~X_L$
$V_L = \frac{\mathscr{E}_m~\omega_d~L}{R}$
$V_L = \frac{\mathscr{E}_m}{R}~\sqrt{\frac{L}{C}}$
$V_L = \frac{10~V}{10~\Omega}~\sqrt{\frac{1.0~H}{1.0\times 10^{-6}~F}}$
$V_L = 1000~V$
The amplitude of the voltage across the inductor is $~~1000~V$
