Find the number of sides of a regular polygon whose interior angle is twice the exterior angle.

Let the measure of each of the interior angles of the polygon be $x$. The corresponding exterior angles will be $\frac{x}{2}$ because the exterior angles are half of the interior angles. The sum of the interior angle and the exterior angle is $180\degree$ so $x +\frac{x}{2} = 180$, $\frac{3x}{2}=180$, $3x=360$, $x=120$. Therefore each of the interior angles of the polygon is $120\degree$.

The angle measure of each of the interior angles of an n-sided regular polygon is $\frac{(n-2)\times180}{n}$. This means that

The polygon referred to in the question has six sides.