Step 1: Check for resonance and find the Q-factor.
Given $X_L = 20\,\Omega$, $X_C = 20\,\Omega$, $R = 10\,\Omega$, and supply $V_s = 200\,V$. Since $X_L = X_C$, the circuit is at resonance. The Q-factor is $Q = X_C / R = 20/10 = 2$.
Step 2: Apply the voltage magnification property.
At resonance, the voltage across the capacitor equals $Q$ times the source voltage: $V_C = Q \times V_s = 2 \times 200 = 400\,V$.
Step 3: Assign the phase angle.
Capacitor voltage always lags the circuit current by $90^\circ$. At resonance, current is in phase with the source voltage. So the capacitor voltage lags the source by $90^\circ$. \[ \boxed{400\angle -90^\circ\,V} \]