Explanation

(a) Power in electric bulb
$P = \frac{V^{2}}{R}$
So, resistance of electric bulb
$R = \frac{V^{2}}{P}$
Given, $P_{1} = 25\ W,\ P_{2} = 100\ W,$
$V_{1} = V_{2} = 220\ volt$
Therefore, for same potential difference $V$
$R \propto \frac{1}{P}$
Thus, we observe that for minimum power, resistance will be maximum and $vice - versa.$
Hence, resistance of 25 W bulb is maximum and 100 W bulb is minimum.