Explanation

(b) Resistance of 40 W bulb $\ = \frac{240\ \times 240}{40}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 1440\ \Omega$
It’s safe current $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{240}{1440} = 0.167\ A$
Resistance of 60 W bulb $= \frac{240\ \times 240}{60}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 960\ \Omega$
It’s safe current $\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = \frac{240}{960} = 0.25\ A$
When connected in series to 420 V supply, then the current
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ I = \frac{420}{1440 + 960} = \frac{420}{2400}$
$\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ = 0.175\ \ A$
Thus, current is greater for 40 W bulb, so it will fuse.