jmc

Let $n$ be the number of sides in the polygon. The sum of the interior angles in any $n$-sided polygon is $180(n-2)$ degrees. Since each angle in the given polygon measures $162^{\circ }$, the sum of the interior angles of this polygon is also $162n$. Therefore, we must have $$180(n-2)=162n.$$ Expanding the left side gives $180n-360=162n$, so $18n=360$ and $n=\boxed{20}$.

We might also have noted that each exterior angle of the given polygon measures $180^{\circ }-162^{\circ }=18^{\circ }$. The exterior angles of a polygon sum to $360^{\circ }$, so there must be $\frac{360^{\circ }}{18^{\circ }}=20$ of them in the polygon.