1. Use the relation between power, voltage, and current

For an electric bulb, $P = VI$ So the current is $I = \frac{P}{V} = \frac{110}{220} = 0.5\ \text{A}$

2. Interpret current in terms of charge flow

Current is charge flowing per second: $I = \frac{Q}{t}$ Since $I = 0.5\ \text{A}$, the charge flowing per second is $Q = 0.5\ \text{C/s}$

3. Find the number of electrons flowing per second

Charge of one electron is $e = 1.6 \times 10^{-19}\ \text{C}$

If $n$ electrons flow per second, then $n = \frac{Q}{e} = \frac{0.5}{1.6 \times 10^{-19}}$

Now calculate: $n = \frac{5}{10} \cdot \frac{1}{1.6 \times 10^{-19}}$ $n = \frac{0.5}{1.6} \times 10^{19}$ $n = 0.3125 \times 10^{19}$ $n = 3.125 \times 10^{18}$

This can also be written as $3.125 \times 10^{18} = 31.25 \times 10^{17}$

4. Match with the options

Option B is: $31.25 \times 10^{17} = 3.125 \times 10^{18}$

So the correct option is B.