Explanation

(c) As for an electric appliance $R = \frac{V^{2}}{P}.$
For first bulb, its resistance
$R_{2} = \frac{V^{2}}{P_{1}} = \frac{250 \times 250}{100} = 625\ \Omega$
For second bulb, its resistance
$R_{2} = \frac{V_{2}^{2}}{P_{2}} = \frac{200 \times 200}{100}$
$\ \ \ \ \ \ = 400\ \Omega$
Now, in series potential divides in proportion to resistance.
So, $V_{2} = \frac{R_{2}}{\left( R_{1} + {\ R}_{2} \right)}\ V$
Where $V$ is supply voltage.
$\therefore$ Potential drop across bulb $B_{2}.$
$V_{2} = \frac{400}{(625 + 400)} \times 250$
$\ \ \ \ \ = 97.56\ V$
$\ \ \ \ \ = 98\ V$