jmc

Let the triangle have vertices $A$, $B$, and $C$, let $O$ be the center of the circle, and let $D$ be the midpoint of $\overline{BC}$. Triangle $COD$ is a $30-60-90$ degree triangle. If $r$ is the radius of the circle, then the sides of $△COD$ are $r$, $\frac{r}{2}$, and $\frac{r\sqrt{3}}{2}$. The perimeter of $△ABC$ is $6\left(\frac{r\sqrt{3}}{2}\right)=3r\sqrt{3}$, and the area of the circle is $\pi r^2$. Thus $3r\sqrt{3}=\pi r^2$, and $r=\boxed{\frac{3\sqrt{3}}{\pi }}$.