Answer
$-9.6^{\circ}$
Work Step by Step
We know that
$X_c=\frac{1}{\omega C}$
$\implies X_c=\frac{1}{2\pi(60)(1.5\mu F)}=1.77K\Omega$
$X_L=\omega L=2\pi(60.0)(250mH)=94.2\Omega$
Now $\phi=tan^{-1}\frac{X_L-X_c}{R}$
We plug in the known values to obtain:
$\phi=tan^{-1}\frac{94.2-1770}{9900}=-9.6^{\circ}$
