Current through the light bulb = 3/2 A = 1.5 A
Assume the transformer is ideal, i.e. with 100% efficiency. By conservation of energy,
power input = power output
220.I = 3
i.e. I = 3/220 A = 0.0136 A = 13.6 mA
The actual current would be larger than 13.6 mA due to various energy losses in the transformer.