imc

The situation is shown in the picture below. The radius we seek is $SD=AD+AS$. Clearly $AD=1$. The point $S$ is the center of the equilateral triangle $ABC$, thus $AS$ is $2/3$ of the altitude of this triangle. We get that $AS=\frac{2}{3}\cdot \sqrt{3}$. Therefore the radius we seek is $1+\frac{2}{3}\cdot \sqrt{3}=\boxed{\frac{3+2\sqrt{3}}{3}}$.