The phasor voltage is
$\begin{align*}\mathbf{V} = \varepsilon_m e^{j \theta}\end{align*}$
where $\varepsilon_m = 40 \mbox{ V}$ . Therefore, the steady state phasor current is
$\begin{align*}\mathbf{I} = \frac{\mathbf{V}}{\mathbf{Z}} = \frac{\varepsilon_m e^{j \theta}}{R - \frac{j}{\omega C} + j \ome...$
The magnitude of the phasor current (which equals the maximum amplitude of sinusoidal current) is
$\begin{align}\left|\mathbf{I} \right| = \frac{\varepsilon_m}{\displaystyle \sqrt{R^2 + \left(\frac{-1}{\omega C} + \omega L \...$
From equation (1), we see that the amplitude of the current is maximized when
$\begin{align*}\frac{-1}{\omega C} + \omega L = 0 \Rightarrow C = \frac{1}{\omega^2 L} = \frac{1}{\left(1000 \mbox{ rad/s}\rig...$
Therefore, answer (D) is correct.

Solution to 2001 Problem 38