Correct Answer  Option 4 : 30
CONCEPT:

Electric Power: The rate at which electrical energy is dissipated into other forms of energy is called electrical power i.e.,
\(P = \frac{W}{t} = VI = {I^2}R = \frac{{{V^2}}}{R}\)
Where V = Potential difference, R = Resistance and I = current.
CALCULATION:
Given  Voltage rating (V) = 30 V, Power rating of the lamp (P) = 90 W
 The resistance of the lamp can be calculated as
\(⇒ R_L= \frac{{{V^2}}}{P}=\frac{(30)^2}{90}=10\, Ω\)
Now, the current in the lamp is
\(⇒ I_L=\frac{V}{R_L}=\frac{30}{10}=3\, A\)
 According to the question, as the lamp is operated on 120V DC, then resistance becomes
\(⇒ R'=\frac{V'}{I_L}=\frac{120}{3}=40\, Ω\)
 For proper glow, a resistor must be connected in series with the lamp, therefore
⇒ R' = R + R_{L}
⇒ R = R'  R_{L} = (40  10) Ω = 30 Ω