Explanation

(d) Bulb (I):Rated current$\ I_{1} = \frac{P}{V} = \frac{40}{220} = \frac{2}{11}amp$
Resistance$\ R_{1} = \frac{V^{2}}{P} = \frac{(220)^{2}}{40} = 1210\Omega$
Bulb (II):Rated current$\ I_{2} = \frac{100}{220} = \frac{5}{11}amp$
Resistance $R_{2} = \frac{(220)^{2}}{100} = 484\ \Omega$
When both are connected in series across $40\ V$ supply
Total current through supply
$I = \frac{40}{P_{1} + P_{2}} = \frac{40}{1210 + 484} = \frac{40}{1254} = 0.03A$
This current is less than the rated current of each bulb. So neither bulb will fuse
Short Trick : Since $V_{Applied} < V_{Rated'}$ neither bulb will fuse