Answer
$\phi = -90^{\circ}$
Work Step by Step
Note that $~~I_{rms} = \frac{\mathscr{E}_{rms}}{Z}$
The average rate at which energy is dissipated in the resistor is a minimum when $I_{rms}$ is a minimum.
This occurs when we maximize the impedance $Z$
$Z = \sqrt{R^2+(X_L-X_C)^2}$
$Z = \sqrt{R^2+(\omega_d~L-1/\omega_d~C)^2}$
Note that when $C = 0$, then $Z \to \infty$ and $I_{rms} = 0$
We can find the phase angle $\phi$ when $C=0$:
$tan~\phi = \frac{X_L-X_C}{R}$
$tan~\phi = \frac{\omega_d~L-1/\omega_d~C}{R}$
$tan~\phi = \frac{\omega_d~L-1/\omega_d~(0)}{R}$
$tan~\phi = -\infty$
$\phi = tan^{-1}~(-\infty)$
$\phi = -90^{\circ}$
