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{align*} 6\pi r^2-12\pi r&=0 \implies \\ 6\pi r( r-2 )&=0 \implies \\ r=0 \quad &\text{or}\quad r=2.\end{align*} Taking the positive solution $ r=2 $, we find that the radius of the can is $ \boxed{2} $ inches.