jmc

A sphere with radius 9 inches has volume $\frac{4}{3}\pi (9^3)=4\cdot 9^2\cdot 3\pi$ cubic inches; twice this is $8\cdot 9^2\cdot 3\pi$ cubic inches. Let the radius of the larger sphere be $r$, so we have $$\frac{4}{3}\pi r^3=8\cdot 9^2\cdot 3\pi .$$ Solving for $r$ yields $$r^3=2\cdot 9^3\Rightarrow r=9\sqrt[3]{2}.$$ The diameter is twice this value, or $18\sqrt[3]{2}$ inches. Hence $a=18$, $b=2$, and $a+b=\boxed{20}$.