Work done by external magnetic field

You need to find the torque $tau$ as a function of $theta$ and then use the work done to rotate the coil through an angle $dtheta$ is $tau ,dtheta$.

If the system is the loop you can say that the direction of the torque on the loop and the direction of the rotation of the loop are the same (equivalent to force and displacement being in the same direction).
So the product of torque and angle of rotation will be positive.

The question is worded incorrectly. This is because the magnetic field is doing no work. (See for instance this page: https://van.physics.illinois.edu/qa/listing.php?id=17176).

You will need to supply an EMF (from a electric power supply) to ensure that the current is constant. Now, that being said, we can rephrase the question one of two ways, 1) what work does the current source do on the loop? or 2) what mechanical work do you need to do to physically twist the loop from $0$ to $60^o$? Question 2 is probably what you meant to ask.

Now you are almost correct in your solution, but recall that the dipole moment of the current loop is perpendicular to the area enclosed by the loop, in the $hat{n}$ dierection. The magnetic dipole moment is given by $mu=IAhat{n}$. The potential energy of a dipole in a magnetic field is $U=-vec{mu}cdotvec{B}$.

So the work you do in rotating the dipole is $W = U_{f}-U_i = [-IAcos (30^o)]-[-IAcos (90^o)]$.

If you plug in the numbers, you will get the answer you are looking for, with a negative sign, which indicates that the loop does work on you. Intuitively, you can think the following way about the sign: magnetic dipoles tend to align with the magnetic fields, so you don't need to do work on the dipole, the dipole does work on you.

So it is negative work for you to physically rotate the dipole from the initial position to the final position. But the current supply has to do the opposite (i.e. positive) amount of work to keep the current flowing.

Hope this clarifies things for you.