Answer
$ \theta \approx0.430^{\circ}$
Work Step by Step
For this problem, we can solve for the angle, $\theta$, using simple right-triangle geometry with the two given distances.
$\sin \theta \approx y/D$
where D is the distance to the screen and y is the distance to the minima.
Therefore;
$ \theta \approx \arcsin y/D$
$ \theta \approx \arcsin 0.0150m/2.00m$
$ \theta \approx0.430^{\circ}$
