We know that P = V2/R
Here V across all the bulbs is 220 V and P = 100 W.
Then the resistance of the bulb is
R = V2/P
= (220)2/100
= 48400/100
= 484 Ω
The circuit when reduce gives 4 resistors each of resistance R = 484 Ω in parallel to each other.
Hence, the equivalent resistance is R/4 = 484/4 Ω = 121 Ω
Total current in the circuit is V/R = 220/121 = 20/11 and this is equally divided among the bulbs.
Current flowing into B2, B3 and B4 will be reading of the ammeter = 20/11 x 3/4 = 15/11 A = 1.36 A