jmc

Let $r$ be the number of inches in the radius of the can. The volume of the can is $\pi r^2(\text{height})=6\pi r^2$ cubic inches. The lateral surface area is $2\pi r(\text{height})=12\pi r$ square inches. Setting $6\pi r^2$ equal to $12\pi r$, we find $$\begin{array}{rl}6\pi r^2-12\pi r& =0⟹\\ 6\pi r(r-2)& =0⟹\\ r=0& \text{or}r=2.\end{array}$$ Taking the positive solution $r=2$, we find that the radius of the can is $\boxed{2}$ inches.