(d) Let the resistance of the lamp filament be $R.$ Then $100 = (220)^{2}/R.$ When the voltage drops, expected power is
$P = (220\ \times 0.3)^{2}/R'$
Here, $R'$ will be less than $R,$ because now the rise in temperature will be less. Therefore, $P$ is more than $(220\ \times 0.9)^{2}\ R = 81\ W.$ But it will not be 90% of earlier value, because fall in temperature is small. Hence, option (d) is correct.