**EDIT:** The isomorphism problem for finitely presented solvable groups in the variety of all solvable groups of derived length $\le 7$ is undecidable. This was proved by Kirkinskiĭ and Remeslennikov (Kirkinskiĭ, A. S.; Remeslennikov, V. N.
`The isomorphism problem for solvable groups.' (Russian)
Mat. Zametki 18 (1975), no. 3, 437–443.). The Russian version of this article can be downloaded from here. The English translation is available here.

Unfortunately this **does not fully answer** the original question, because the groups in this construction are finitely presented in the variety of solvable groups but may not be finitely presented in the variety of all groups.

I would guess that one could use O. Kharlampovich's example of a finitely presented 3-step solvable group with unsolvable word problem (Harlampovič, O. G., `A finitely presented solvable group with unsolvable word problem.' (Russian) Izv. Akad. Nauk SSSR Ser. Mat. 45 (1981), no. 4, 852–873, 928.) to construct the family of groups you need. Perhaps someone has already done this...

**Second EDIT:** Indeed, this was done by Baumslag, Gildenhuys and Strebel (see Theorem 1 in Baumslag, Gilbert; Gildenhuys, Dion; Strebel, Ralph, `Algorithmically insoluble problems about finitely presented solvable groups, Lie and associative algebras. II.'
J. Algebra 97 (1985), no. 1, 278–285.), who proved that the isomorphism problem is undecidable in the class of finitely presented solvable groups of derived length 3.

In fact, in one of her talks Olga Kharlampovich mentioned that she can construct a finitely presented 3-step solvable group $G$ with unsolvable word problem that is Hopfian. Then the isomorphism problem among the quotients of $G$ by one defining relation is unsolvable (because $G/\langle\langle g \rangle\rangle^G$ is isomorphic to $G$ if and only if $g=1$ in $G$).