Answer
Option d.
The loop is pushed to the right, out of the magnetic field.
Work Step by Step
By Faraday's Law, $\displaystyle \varepsilon = -\frac{d\Phi_B}{dt} = -(\vec{A}\frac{\vec{B}}{dt} + \vec{B}\frac{d\vec{A}}{dt})$, the induced emf is dependent on a changing magnetic flux, or a changing cross sectional area and magnetic field. In this problem, the cross sectional area is not changing right away, so we can say that $\qquad\varepsilon \propto \displaystyle -\frac{d\vec{B}}{dt}$.
Thus, we can imagine that there is an induced magnetic field in the opposite direction of the regular magnetic field (the regular magnetic field in the question is in the negative $z$-direction (into the page; ($-\hat k$)); the "induced" magnetic field we're imagining is then in the positive $z$-direction, out of the page). Using the right-hand curl, with our thumb pointing towards us, we see that the induced current flows counterclockwise throughout the loop.
The magnetic force points in three different directions because there's three different parts of the loop in the magnetic field: the top horizontal part (where the induced current flows to the left), a single vertical part (where the induced current flows downward), and the bottom horizontal part (where the induced current flows rightward). The net magnetic force on the horizontal parts of the loop will cancel out since they're going in opposite directions, so we only need to be concerned with the vertical part with a downward induced current flowing through it.
The induced current flows in the ($-\hat j$) direction (downward), and the given, regular magnetic field in the question still flows into the page in the ($-\hat k$) direction. The direction of the magnetic force on the vertical piece of the loop inside the magnetic field is $\vec{F}_B = I\vec{L} \times \vec{B} = (-\hat j) \times (-\hat k) = +\hat i$
Thus, the direction of the magnetic force is the positive $x$-direction, or to the right. Option d. is correct.
