Correct Answer - `(sqrt(3))/(pi)H`
From the rating of the bulb, the resistance of the blub is
`R=(V^2)/(P) = 100 Omega`
For the blub to be operated at its rated value the rms current through it should be 1A.
Also, `I_(rms)=(V_(rms))/(Z)` ltbr gt`:. 1=(200)/(sqrt(100^(2)+(2 pi 50 L)^(2)) implies L= (sqrt(3))/(pi) H`.