Answer
We can rank the loops according to the size of the current induced in them:
$2 \gt 1 = 3$
Work Step by Step
$\Phi = B~A$
$\mathscr{E} = -\frac{d\Phi}{dt} = -A~\frac{dB}{dt}$
The current in the long wire produces a magnetic field around the wire. Since the current is increasing, the magnitude of the magnetic field also increases.
By the right hand rule, the magnetic field is directed out of the page above the wire, and into the page below the wire.
By symmetry in loop 1 and loop 3, this changing magnetic field produces a changing flux through the top half of the loop, and a changing flux through the bottom half of the loop in the opposite direction. The net change in magnetic flux is zero. Therefore, there is no change in magnetic flux through loop 1 and loop 3 so there is no induced emf and no induced current.
Since loop 2 is not symmetrical about the long wire, the changing magnetic field produces a non-zero change in magnetic flux through loop 2. Therefore, there will be an induced emf and an induced current.
We can rank the loops according to the size of the current induced in them:
$2 \gt 1 = 3$
