The answer lies in Lemma 2.1 of Sullivan, Curves of Finite Total Curvature.
The paper refers to exterior angles as turning angles, and for a polygon $P$ designates the sum of turning angles (total curvature) as $TC(P)$
Here's a paraphrased version of the key lemma:
Lemma 2.1. Suppose $P$ is a polygon in $\mathbb R^3$. If $P'$ is obtained from $P$ by deleting one vertex $v_n$ then $TC(P') ≤ TC(P)$.
We have equality here if $v_{n−1}v_n v_{n+1}$ are collinear in that
order, or if $v_{n−2}v_{n−1}v_{n}v_{n+1}v_{n+2}$ lie convexly in some
plane, but never otherwise.
Please refer to the paper for a proof, along with background and definitions. It's a short simple argument, using a mapping that takes 3D polygons $P$ to spherical polygons $P_S$ on the unit sphere $\mathbb S^2$. The total curvature of $P$ is the same as the boundary length of $P_S$.
Lemma 2.1 nicely takes care of both cases.
For a non-planar polygon $P$, removing vertices until only a triangle $T$ is left defines a sequence of polygons whose total curvature is strictly decreasing in at least one pair of the sequence (corresponding to where some $v_{n−2}v_{n−1}v_{n}v_{n+1}v_{n+2}$ is not planar). Since $TC(T)=2\pi$ we conclude that $TC(P)>2\pi$.
Similarly, for complex planar polygons $P$ there will be vertex
sequences $v_{n−2}v_{n−1}v_{n}v_{n+1}v_{n+2}$ that are not convex and again $TC(P)>2\pi$.
Best Answer
The sum of all exterior angles in any polygon is $\mathbf{360^{\circ}}$. Thus, we have all the divisors of $360$ as a possible exterior angle as a choice and the number we have to divide $360$ by to get the divisor will be the number of sides in the particular polygon.
For example, we have $90$ as a divisor of $360$ so $90^{\circ}$ is a possible exterior angle. Since $90 \times 4 = 360$, the particular polygon with $90^{\circ}$ exterior angle will have $4$ sides.
Another fact: the sum of angles in a polygon with $n$ sides is $180(n - 2)$ degrees. So all the sums will be a multiple of $180$. Use that formula with $n = 8$ to get the sum of all interior angles in an octagon.
Remark: Technically, all the choices you have been provided with can be the measures of an exterior angle. If in the future, any question asks about regular polygons, you can predict that the answers will only be the divisors of $360$.