jmc

In order to make the glass sphere as small as possible, the plum and the watermelon should be touching---that is, they should be externally tangent spheres. Since the plum has a point that has distance 20 from another point on the watermelon, any sphere containing the plum and the watermelon must have radius at least 10. On the other hand, Rose can enclose them both in a sphere of radius 10, as shown in the diagram below:



Thus the smallest sphere that can contain the plum and the watermelon has radius 10. So it remains to subtract the volumes of a sphere of radius 2 and a sphere of radius 8 from a sphere of radius 10. Since the volume of a sphere of radius $r$ is $\frac{4}{3}\pi r^3$, it follows that the volume in question is $$\begin{array}{rl}\frac{4}{3}\pi \cdot 10^3-\frac{4}{3}\pi \cdot 8^3-\frac{4}{3}\pi \cdot 2^3& =\frac{4}{3}\pi (10^3-8^3-2^3)\\ & =\frac{4}{3}\pi (1000-512-8)\\ & =\frac{4}{3}\pi \cdot 480=640\pi .\end{array}$$Therefore our answer is $\boxed{640}$.

We could also have simplified the final computation by noting that in general $$(a+b{)}^3-a^3-b^3=3a^{2}b+3a{b}^2=3ab(a+b).$$Setting $a=2$ and $b=8$, we have $$\begin{array}{rl}\frac{4}{3}\pi (a+b{)}^3-\frac{4}{3}\pi a^3-\frac{4}{3}\pi b^3& =\frac{4}{3}\pi [(a+b{)}^3-a^3-b^3]\\ & =\frac{4}{3}\pi \cdot 3ab(a+b)=4\pi ab(a+b).\end{array}$$This tells us that $K=4ab(a+b)=4\cdot 2\cdot 8\cdot 10=640$, as before.