Answer
$v_c=11.9 V$
Work Step by Step
The capacitor voltage phasor is $90^{\circ}$ less than that of the current. Thus, at $t=4.17 \mathrm{~ms}$, we obtain
$
\begin{aligned}
v_C & =I \sin \left(90^{\circ}-\left(-24.3^{\circ}\right)-90^{\circ}\right) X_C\\&=I X_C \sin \left(24.3^{\circ}\right)=(0.164 \mathrm{~A})(177 \Omega) \sin \left(24.3^{\circ}\right) \\
& =11.9 \mathrm{~V} .
\end{aligned}
$
