#### Answer

a.) $1 .3 \cdot 10^{-11}$ N$\cdot$m
b.) The loop will flip 90 degrees clockwise

#### Work Step by Step

The torque on the current loop can be represented as $\vec{\tau} = \vec{m} \times \vec{B}$ where $\vec{m} = I\vec{A}$ is the magnetic moment. The torque on the current loop is
$\vec{\tau} = \vec{m} \times \vec{B} = \vec{m}\vec{B}\sin(\theta) = (I\vec{A})\vec{B}\sin(\theta)$
But which current do we use when calculating the torque? What is the magnetic field?
The magnetic field is of the wire is $\displaystyle \vec{B} = \frac{\mu_0I_{wire}}{2\pi d}$ and we'll use the current in the square loop to calculate the torque since that's the current at the point in space where the loop is. Keep in mind, the question said $loop$ and not $square$ $loop$, so $\vec{A} = \pi r^2$ . Also, the magnetic moment of the loop (which is in the left/rightward and into/out of the page directions, is completely perpendicular to the magnetic field generated by the wire that's flowing in the upward/downward directions at the loop (use the right-hard curl rule)).
This means
$\vec{\tau} = I\vec{A}\vec{B}\sin(\theta) = \displaystyle I_{loop}(\pi r^2)\cdot \frac{\mu_0I_{wire}}{2\pi d} \sin(90) $
$\vec{\tau} \displaystyle = (0.20)(\pi (\frac{2\cdot10^{-3}}{2})^2)\cdot\frac{(4\pi \cdot10^{-7})(2.0)}{2\pi (0.020)}(1)$
$\vec{\tau} \displaystyle = (0.20)(\pi(10^{-3})^2)\frac{4\cdot10^{-7}}{0.020}$
$\vec{\tau} = 1.3 \cdot 10^{-11}$ N$\cdot$m
b.) Recall that two current-carrying wires attract each other through magnetic force when their currents flow in the same direction, thus when the wires have currents flowing in the opposite direction there is a repulsive magnetic force on each wire.
Looking at the figure provided, we see that the bottom of the current loop has current flowing in the same direction as the wire (both are out of the page) and the top of the current loop has current flowing in the opposite direction of the wire (the current flowing in the top of the loop is into the page). Thus, there will be an attractive force on the bottom part of the current loop and a repulsive force on the top part of the current loop. The loop will flip 90 degrees so that the part of the loop with the current flowing in the same direction is closer to the current-carrying wire than the part of the loop that doesn't have current flowing in the same direction.
Thus, the loop flips 90 degrees clockwise.