An electric bulb is rated $220\text{V}-100\text{W}$. The power consumed by it when operated on $110\text{V}$ will be:

Resistance of the bulb $R=\frac{V_{rated}^2}{P_{rated}}=\frac{220^2}{100}=484\Omega$.Power consumed at $110\text{V}$:
$$P=\frac{V^2}{R}=\frac{110^2}{484}=\frac{12100}{484}=25\text{W}$$
Alternative method: $P\propto V^2$ for constant $R$.If voltage is halved ($220\to 110$), power becomes $(1/2{)}^2=1/4$ of original.$P=100/4=25\text{W}$.