jmc

The circumference of the whole circle is $2\pi \cdot 6=12\pi$, so the circumference of the base of the cone is $12\pi /3=4\pi$. Hence, the radius of the base of the cone is $4\pi /(2\pi )=2$.



The slant height of the cone is 6 (the radius of the original circle), so by Pythagoras, the height of the cone is $\sqrt{6^2-2^2}=\sqrt{32}=\boxed{4\sqrt{2}}$.