smc

By Vieta's formulas, the sum of the six roots is $10$ and the product of the six roots is $16$. By inspection, we see the roots are $1,1,2,2,2,$ and $2$, so the function is $(z-1{)}^2(z-2{)}^4=(z^2-2z+1)(z^4-8z^3+24z^2-32z+16)$. Therefore, calculating just the $z^3$ terms, we get $B=-32-48-8=\boxed{-88}$.