Correct Answer - A
Power of electric bulb
`P=(V^(2))/R`
So, resistance of electric bulb `R=(V^(2))/P`
Given `P_(1)=25 W , P_(2)=100 W, V_(1)=V_(2)=200V`
Therefore, for same potential difference `V, R prop 1/P`
Thus, we observe that for minimum power, resistance will be maximum and vice-versa. hence, resistance of 25W bulb is maximum and 100 W bulb is minimum.