Chapter 4 problem 24.33
Wire 1 has an attracting force from wire 3 at 0.04 m distance and a repelling force from wire 2 at 0.02 m distance.
F = µ*I1*I2*L/(2πd)
When d = 0.02 m, F = 0.0005 N
When d = 0.04 m, F = 0.00025 N
Net force on wire 1 is 0.00025 N repelling, or upward.
By symmetry the net force on wire 3 is the same but downward.
Wire 2 has equal repelling forces from wires 1 and 3 so net force = 0.
For an explanation of the forces, see the ref.