jmc

We draw right triangle $GAC$ within the cube below:



Since $\overline{AG}$ is a space diagonal of the cube, we have $AG=CG\cdot \sqrt{3}$. Therefore, considering right triangle $AGC$ gives us $$\sin \angle GAC=\frac{CG}{AG}=\frac{CG}{(\sqrt{3})(CG)}=\frac{1}{\sqrt{3}}=\boxed{\frac{\sqrt{3}}{3}}.$$