Explanation

(c) Resistance of electric bulb $R = \frac{V^{2}}{P}$ where subscripts denote for rated parameters.
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ R = \frac{(220)^{2}}{100}$
Power consumed at 110 V, $P_{consumed} = \frac{V^{2}}{R}$
$\therefore\ \ P_{consumed} = \frac{(110)^{2}}{(220)^{2}/100} = 25\ W$