1. Given data

• Inductance: $L = 200\,\text{mH} = 0.2\,\text{H}$
• AC source rms voltage: $V_{\text{rms}} = 220\,\text{V}$
• Frequency: $f = 50\,\text{Hz}$

We need the peak current when the instantaneous source voltage is zero.

2. Current in a pure inductor

For a pure inductor, current lags voltage by $\dfrac{\pi}{2}$.

If $v = V_0 \sin \omega t,$ then $i = I_0 \sin\left(\omega t - \frac{\pi}{2}\right).$

When the instantaneous voltage is zero, $v=0 \implies \sin \omega t = 0.$ At that instant, the current magnitude is maximum, i.e. equal to peak current $I_0$.

So the asked current is simply the peak current.

3. Inductive reactance

$X_L = \omega L = 2\pi f L$

Substitute values: $X_L = 2\pi \cdot 50 \cdot 0.2 = 20\pi\,\Omega$

4. RMS current

$I_{\text{rms}} = \frac{V_{\text{rms}}}{X_L} = \frac{220}{20\pi} = \frac{11}{\pi}\,\text{A}$

5. Peak current

$I_0 = \sqrt{2}\, I_{\text{rms}} = \sqrt{2}\cdot \frac{11}{\pi} = \frac{11\sqrt{2}}{\pi}$

Given that this peak current is $\frac{\sqrt{a}}{\pi}\,\text{A}$

So, $\sqrt{a} = 11\sqrt{2}$

Squaring both sides, $a = 121 \times 2 = 242$

6. Final answer

$\boxed{242}$