Refer to the figure given below, current between terminals $$A$$ and $$B$$ is _______A.

Look at one horizontal branch (top/middle/bottom):

each branch has
3 cells of 5 V and 3 resistors of 3 Ω in series

so for each branch:
net emf = 5 + 5 + 5 = 15 V
total resistance = 3 + 3 + 3 = 9 Ω

so current in each such branch (if isolated):

I = 15 / 9 = 5/3 A

Now there are 3 identical branches in parallel, all between the same left and right nodes

so total current from these three:

$$I_{total}(top\ part)=3\times(5/3)=5A$$

bottom branch (A to B):

only resistors: 3 + 3 + 3 = 9 Ω
no battery

so it behaves like a resistor connected across the same two nodes

Now the three upper branches together act like a single source supplying 5 A between the nodes

this current must pass through the bottom 9 Ω path

so:

I = 5 A