Let $G$ be an hypercentral group such that $G/G'$ is a torsion factor group. Is it true that then $G$ also is a torsion group?
If it were nilpotent then it would be easy by a result of Robinson, but now? I think it is true but i do not know how to prove it.
Notation
A group is called hypercentral if it possess a central ascending series.
I don't know much about hypercentral groups, but it seems to be well known that they are locally nilpotent, which implies that their torsion elements form a normal subgroup. So, to prove your result, we can factor out the torsion subgroup and assume that $G$ is torsion-free.
If $G$ is abelian then it is trivial and we are done, so suppose not, and let $h$ be a fixed element in $Z_2(G) \setminus Z(G)$. Then the map $\phi:G \to Z(G)$ with $g \mapsto [h,g]$ is a nontrivial homomorphism onto a torsion-free abelian group, so $G/\ker(\phi)$ is abelian but not torsion-free, and hence $G/G'$ cannot be torsion-free.