finitely presented subgroup and free solvable group of class 3 Let $F(n)$ be free group of rank $n\geq 2$.  Denote  by $F_d(n)$ the d-th derived subgroup, that is $F_d(n)=[F_{d-1}(n),F_{d-1}(n)]$ where $F_0(n)=F(n)$. The free solvable group of rank $n$ and solvable class $d$ is $S_{n,d}=F(n)/F_d(n)$.
By Corollary 2.14. in "FINITELY PRESENTED WREATH PRODUCTS AND DOUBLE
COSET DECOMPOSITIONS",(http://arxiv.org/pdf/math/0509090.pdf), the free solvable group of class 3 and finite rank  is not a finitely presented group. Does this group contain a finitely presented subgroup of solvable class 3?
 A: No, it does not exist.
It's not too hard to check that a fp subgroup of a free metabelian group is abelian. A consequence (since any quotient of a fp solvable group is fp, by Bieri-Strebel) is that any fp subgroup of a free 3-solvable group has abelian image in the metabelianization, and in particular, has to be metabelian (I'd guess that it even has to be abelian!).
I have to justify the first claim (fp subgroup of a free metabelian group $G$ is abelian). Write, $M=[G,G]$, $Q=G/M$. Then $M$ is a torsion-free $\mathbf{Z}Q$-module [edit: justification below].
Let $H$ a fg subgroup of $G$, and $Q_H$ be the image of $H$ in $Q$.
If If $M\cap H$ or $Q_H$ is trivial then clearly $H$ is abelian.
So assume otherwise, and let us show that $H$ is not finitely presented.
Then $M\cap H$ is also, by restriction, a torsion-free $\mathbf{Z}Q_H$-module. So it is not finitely-generated over the subsemigroup $\{q\in Q_H:\ell(q)\ge 0\}$ for any choice of nonzero homomorphism $Q_H\to\mathbf{R}$. This means that the Bieri-Strebel invariant of the $Q_H$-module $M\cap H$ is the whole space of homomorphism (in conventional literature, it will be rather called empty Bieri-Strebel invariant: I always define the Bieri-Strbel invariant as the closed complement rather than the open subset, in accordance to the idea that we define the spectrum of a matrix rather than the complement of its spectrum). Since a f.g. metabelian group $G$ is f.p. iff its Bieri-Strebel invariant (a closed subset of $\mathrm{Hom}(G,\mathbf{R})$, closed under positive scalar multiplication) contains a line, we deduce that $H$ is not finitely presented.

[Edit:] That $M$ is a torsion-free $\mathbf{Z}Q$-module is a consequence of Magnus' embedding theorem. Assume, as we can, that $G$ is finitely generated free metabelian over generators $x_1,\dots,x_n$. Let $t_1,\dots,t_n,e_1,\dots,e_n$ be complex numbers, such that $(t_1,\dots,t_n)$ is algebraically independent over $\mathbf{Q}$, and $(e_1,\dots,e_n)$ are linearly independent over $\mathbf{Q}(t_1,\dots,t_n)$ (of course this holds if $(t_1,\dots,t_n,e_1,\dots,e_n)$ is algebraically independent).
Magnus says that mapping $x_i\mapsto\begin{pmatrix}t_i & e_i\\ 0 & 1\end{pmatrix}$ yields an injective homomorphism from $G$ to $\mathrm{GL}_2(\mathbf{C})$. Actually, it maps into the subgroup $H$ consisting of elements of the form $\begin{pmatrix}t & v\\ 0 & 1\end{pmatrix}$, where $t$ has the form $\prod t_i^{m_i}$, $(m_1,\dots,m_n)\in\mathbf{Z}^n$, and $v$ belongs to the $\mathbf{Z}[t_1^{\pm 1},\dots,t_n^{\pm 1}]$-submodule generated by $(e_1,\dots,e_n)$. We see $H$ is just a standard wreath product $$\mathbf{Z}^n\wr\mathbf{Z}^n=\mathbf{Z}[t_1^{\pm 1},\dots,t_n^{\pm 1}]^n\rtimes\mathbf{Z}^n=(\mathbf{Z}Q)^n\rtimes Q.$$ In particular, $M$ is isomorphic to a $\mathbf{Z}Q$-submodule of a free $\mathbf{Z}Q$-module of rank $n$, so is torsion-free. 
