smc

Consider a trapezoid (label it $ABCD$ as follows) cross-section of the truncate cone along a diameter of the bases: Above, $E,F,$ and $G$ are points of tangency. By the Two Tangent Theorem, $BF=BE=18$ and $CF=CG=2$, so $BC=20$. We draw $H$ such that it is the foot of the altitude $\overline{HD}$ to $\overline{AB}$: By the Pythagorean Theorem, $$r=\frac{DH}{2}=\frac{\sqrt{20^2-16^2}}{2}=12$$ Therefore, the answer is $\boxed{6}.$