jmc

Let $a$ be the inverse of $201$ modulo $299$. Then, by the definition of the inverse, $201\cdot a\equiv 1(\mathrm{mod}299)$. We are looking for an integer $a$ that satisfies this congruence.

To make our task easier, we note that $603\equiv 5(\mathrm{mod}299)$, and so $$\begin{array}{rl}603\cdot 60& \equiv 5\cdot 60\\ & =300\\ & \equiv 1(\mathrm{mod}299).\end{array}$$Now we write $603$ as $201\cdot 3$: $$201\cdot 3\cdot 60\equiv 1(\mathrm{mod}299).$$Thus, the inverse we seek is $a=3\cdot 60=\boxed{180}$.