The question states that a house is protected by a 9A fuse and is supplied with 220V. We need to determine the maximum number of 60 watt bulbs that can be turned on in parallel.

To calculate the maximum number of bulbs, we need to consider the power consumed by each bulb and the maximum current allowed by the fuse.

Given that each bulb is 60 watts and the supply voltage is 220V, we can use the formula P=IV, where P is power, I is current, and V is voltage. Rearranging the formula, we get I=P/V.

For a 60 watt bulb, the current consumption is I=60W/220V = 0.27A.

Since the fuse is rated at 9A, the maximum current allowed is 9A.

To find the maximum number of bulbs that can be turned on in parallel, we divide the maximum current allowed by the current consumed by each bulb: 9A/0.27A = 33 bulbs.

Therefore, the correct answer is option 3: 33.