jmc

Let $\overline{AB}$ and $\overline{DC}$ be parallel diameters of the bottom and top bases, respectively. A great circle of the sphere is tangent to all four sides of trapezoid $ABCD$. Let $E,F$, and $G$ be the points of tangency on $\overline{AB}$, $\overline{BC}$, and $\overline{CD}$, respectively. Then $$FB=EB=18\text{and}FC=GC=2,$$so $BC=20$. If $H$ is on $\overline{AB}$ such that $\angle CHB$ is a right angle, then $HB=18-2=16.$ Thus $$CH=\sqrt{20^2-16^2}=12,$$and the radius of the sphere is $(1/2)(12)=\boxed{6}$.