Answer
$I_1=5.875A$, towards the right
Work Step by Step
We know that
$B_{Net}=B_1-B_2$
We plug in the known values to obtain:
$2.5\times 10^{-6}=\frac{\mu_{\circ}I_1}{2\pi r}-\frac{\mu_{\circ}I_2}{2\pi r}$
$\implies 2.5\times 10^{-6}=\frac{\mu_{\circ}}{2\pi r}(I_1-I_2)$
$\implies \frac{4\pi \times 10^{-7}}{2\pi \times 0.11}(I_1-4.5)=2.5\times 10^{-6}$
This simplifies to:
$I_1=5.875A$, towards the right.
