To find the peak current ($i_m$), we first calculate the peak voltage ($V_m$) and the capacitive reactance ($X_C$).

1. Peak Voltage ($V_m$): The given source voltage is the RMS value ($V_{rms} = 210$ V). According to the sources, $V_m = V_{rms}\sqrt{2} = 210 \times 1.414 \approx 296.94$ V .
2. Capacitive Reactance ($X_C$): $X_C = \frac{1}{\omega C} = \frac{1}{2\pi f C}$. Substituting the values: $X_C = \frac{1}{2 \times 3.14 \times 50 \times 10 \times 10^{-6}} = \frac{1}{314 \times 10^{-5}} \approx 318.47 \ \Omega$ .
3. Peak Current ($i_m$): $i_m = \frac{V_m}{X_C} = V_m \omega C = 296.94 \times (314 \times 10^{-5}) \approx 0.932$ A. Thus, the peak current is nearly 0.93 A.