Let R be the resistance of total length of circular wire then R=36Ω.
The resistance of smaller arc AB R1=R/6=36/6=6Ω
The resistance of bigger arc AB R2=5R/6=5×36/6=30Ω
Now, R1 and R2 are in parallel in between points A and B.
So, effective resistance between A and B =6+306×30=5Ω