Answer
The magnitude of the net magnetic force per unit length on wire 3 due to the currents in the other wires is $~~8.00\times 10^{-7}~N/m$
Work Step by Step
We can write the general expression for the magnetic force on a current-carrying wire due to a parallel current-carrying wire:
$F = \frac{\mu_0~L~i_1~i_2}{2\pi~d}$
Note that currents in the same direction attract each other.
The force on wire 3 due to wire 1 is directed in the -y direction.
The force on wire 3 due to wire 2, wire 4, and wire 5 is directed in the +y direction.
We can find the magnitude of the net magnetic force per unit length on wire 3 due to the currents in the other wires:
$F = -\frac{\mu_0~L~i_1~i_3}{(2\pi)~(2d)}+\frac{\mu_0~L~i_2~i_3}{(2\pi)~(d)}+\frac{\mu_0~L~i_3~i_4}{(2\pi)~(d)}+\frac{\mu_0~L~i_3~i_5}{(2\pi)~(2d)}$
$\frac{F}{L} = \frac{\mu_0~i_2~i_3}{(2\pi)~(d)}+\frac{\mu_0~i_3~i_4}{(2\pi)~(d)}$
$\frac{F}{L} = \frac{2\mu_0~i_2~i_3}{(2\pi)~(d)}$
$\frac{F}{L} = \frac{(2)(4\pi\times 10^{-7}~H/m)(4.00~A)(0.250~A)}{(2\pi)(0.500~m)}$
$\frac{F}{L} = 8.00\times 10^{-7}~N/m$
The magnitude of the net magnetic force per unit length on wire 3 due to the currents in the other wires is $~~8.00\times 10^{-7}~N/m$
