1. Use the bulb rating to find the rms current

For a bulb rated $100\,\text{W}-220\,\text{V}$, when connected to $220\,\text{V}$ rms supply:

$P = V_{\text{rms}} I_{\text{rms}}$

So,

$I_{\text{rms}} = \frac{P}{V_{\text{rms}}} = \frac{100}{220} = 0.4545\,\text{A}$

2. Relate peak current and rms current

For alternating current,

$I_0 = \sqrt{2}\, I_{\text{rms}}$

Thus,

$I_0 = \sqrt{2} \times 0.4545 \approx 1.414 \times 0.4545 \approx 0.643\,\text{A}$

3. Match with the options

$I_0 \approx 0.64\,\text{A}$

So the correct option is B.