Answer
The net magnetic force on wire 2 due to the currents in the other wires is $~~(1.88\times 10^{-4}~N)~\hat{j}$
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 2 due to the current in wire 1 is directed in the -y direction.
The force on wire 2 due to the currents in wire 3, wire 4, and wire 5 is directed in the +y direction.
We can find the magnetic force on wire 2 due to the other wires:
$F = -\frac{\mu_0~L~i^2}{2\pi~d}+\frac{\mu_0~L~i^2}{2\pi~d}+\frac{\mu_0~L~i^2}{(2\pi)~(2d)}+\frac{\mu_0~L~i^2}{(2\pi)~(3d)}$
$F = \frac{\mu_0~L~i^2}{(2\pi)~(2d)}+\frac{\mu_0~L~i^2}{(2\pi)~(3d)}$
$F = \frac{3~\mu_0~L~i^2}{(2\pi)~(6d)}+\frac{2~\mu_0~L~i^2}{(2\pi)~(6d)}$
$F = \frac{5~\mu_0~L~i^2}{(2\pi)~(6d)}$
$F = \frac{(5)(4\pi\times 10^{-7}~H/m)(10.0~m)(3.00~A)^2}{(2\pi)(6)(0.0800~m)}$
$F = 1.88\times 10^{-4}~N$
The net magnetic force on wire 2 due to the currents in the other wires is $~~(1.88\times 10^{-4}~N)~\hat{j}$
