Explanation

(d) Resistance of the lamp
$R_{0} = \frac{V^{2}}{P} = \frac{(30)^{2}}{90} = 10\ \Omega$
Current in the lamp
$i = \frac{30}{10} = 3\ \Omega$
When lamp is operated on a 120 V, then resistance
$R^{'} = \frac{V^{'}}{i} = \frac{120}{30} = 40\ \Omega$
Thus, for proper glow, the resistance required to the put in series will be
$R = R^{'} - R_{0} = 40 - 10 = 30\ \Omega$