$A=\{2,-1+i\sqrt3,-1-i\sqrt3\}$ (cube roots of $8$). Factoring $z^3-8z^2-8z+64=(z-8)(z^2-8)$ gives $B=\{8,2\sqrt2,-2\sqrt2\}$.
- All points of $A$ lie on the circle $|z|=2$, so for $b\in B$ the farthest point of $A$ is at distance at most $|b|+2$. For $b=\pm2\sqrt2$ this is at most $2\sqrt2+2<5$, while the distances from $8$ are larger: $|8-2|=6$ and $|8-(-1\pm i\sqrt3)|$.
- So the greatest distance is between $8$ and $-1+i\sqrt3$ (or its conjugate): $$\left|(-1-8)+i\sqrt3\right|=\sqrt{81+3}=\sqrt{84}=\boxed{2\sqrt{21}}.$$
- Answer is (D).
