Answer
(a) repulsive
(b) $5.7\mu N$
Work Step by Step
(a) We know that the field due to the current in the wire is into the page and thus the current in wire 2 should be flowing from left to right. We also know that when parallel wires carry currents in opposite directions, then the force between the wire is repulsive.
(b) We know that
$F=\frac{I_1Lr_2}{d}(B_{total}+\frac{\mu_{\circ}I}{2\pi r_1})$
We plug in the known values to obtain:
$F=\frac{3.73A\times0.71m\times 0.11m }{0.22m}[2.1\times 10^{-6}+\frac{2\times 10^{-7}\times 3.7A}{0.33m}]$
$F=5.7\times 10^{-6N}=5.7\mu N$
