Explanation:

- The bulb is rated at 100 W for a 110 V supply.

- Power (P) is calculated as P = V²/R.

- For 100 W at 110 V, the resistance (R) can be calculated as:

$$

R = \frac{V^2}{P} = \frac{110^2}{100} = 121 \ \Omega

$$

- When connected to a 220 V power supply, the new power (P') is:

$$

P' = \frac{220^2}{R} = \frac{220^2}{121} \ \approx 400 \ W

$$

- Hence, Option 4: 400 W is the correct answer.