Answer
(a) toward the wire
(b) $3\times 10^{-5}N$
Work Step by Step
(a) We know that the direction of the force on each side of the loop is away from the center of the loop because the force gets stronger closer to the wire. Hence, the net force acts towards the wire.
(b) We know that
$F_{net}=IL\frac{\mu_{\circ}I^{\prime}}{2\pi}(\frac{1}{r_{close}}-\frac{1}{r_{far}})$
We plug in the known values to obtain:
$F_{net}=(2.5A)(1.0m)\frac{(4\pi \times 10^{-7}\frac{T.m}{A})(14A)}{2\pi}(\frac{1}{0.2m}-\frac{1}{1.2m})$
$F_{net}=3\times 10^{-5}N$
