Answer: A

Solution:

By Vieta's formulas the six roots sum to $10$ (the negative of the $z^5$ coefficient) and multiply to $16.$ Thus every root is a power $2^{e_i},$ with $\sum e_i=4.$ Since $2^{e_i}\ge 1+e_i,$ the roots have sum at least $6+4=10.$ Equality holds only when every $e_i$ is $0$ or $1,$ so the roots must be $2,2,2,2,1,1.$

So the polynomial is $(z - 1)^2 (z - 2)^4.$ Expanding, $$\begin{aligned} &(z^2 - 2z + 1) \\ &\quad {}\cdot (z^4 - 8z^3 + 24z^2 - 32z + 16) \\ &= z^6 - 10z^5 + 41z^4 \\ &\quad {}- 88z^3 + 104z^2 \\ &\quad {}- 64z + 16. \end{aligned}$$ The coefficient of $z^3$ is $B = -88.$

Thus, the correct answer is A.