Answer
The current in wire 2 is $~~4.3~A$
Work Step by Step
Note that the distance of wire 1 from point $P$ is $3~d_1$
Note that the distance of wire 2 from point $P$ is $2~d_1$
We can write the general expression for the magnetic field produced by a current in a straight wire:
$B = \frac{\mu_0~i}{2~\pi~R}$
If the net magnetic field at point $P$ is zero, the magnetic field due to each current must be equal in magnitude and opposite in direction.
To find the current $i_2$, we can equate the magnitude of the magnetic field due to each current:
$\frac{\mu_0~i_2}{(2~\pi)~(2~d_1)} = \frac{\mu_0~i_1}{(2~\pi)~(3~d_1)}$
$\frac{i_2}{2} = \frac{i_1}{3}$
$i_2 = \frac{2~i_1}{3}$
$i_2 = \frac{(2)~(6.5~A)}{3}$
$i_2 = 4.3~A$
The current in wire 2 is $~~4.3~A$
