Update:
My memory was quite blurry about this when I originally answered.
See Gonzáles-Acuña, Gordon, Simon, ``Unsolvable problems about higher-dimensional knots and related groups,'' L’Enseignement Mathématique (2) 56 (2010), 143-171.
They prove that any finitely presented group is a subgroup of the fundamental group of the complement of a closed orientable surface in the $4$-sphere, which is much better than I reported.
Original answer:
You most likely would like a finitely presented group, but this might be of interest anyway:
Let $S$ be a recursively enumerable non-recursive subset of the natural numbers and consider the group
$\langle \ a,b,c,d \ | \ a^iba^{-i} = c^idc^{-i} \ \mathrm{for}\ i \in S \rangle$
This has unsolvable word problem. See page 110 of Chiswell's book "A course in formal languages, automata and groups" available on google books (I think it's also in Baumslag's "Topics in Combinatorial Group Theory" but all my books are in boxes at the moment.)
This should be the fundamental group of the complement of a noncompact surface in $\mathbb{R}^4$. You do this in the usual way by beginning with the trivial link on four components and then drawing the movie of the surface in $\mathbb{R}^4$, band summing at each stage to make the conjugates of $b$ and $d$ equal.
I think you end up with a knotted disjoint union of planes. I remember doing this at some point in graduate school when C. Gordon asked me if there were any compact surfaces in the $4$-sphere whose complements have groups with unsolvable word problem.
Best Answer
One general method is to consider an infinite presentation of the group, and then show that every finite subset of the set of relations defines a group with clearly different property. for example, the lamplighter group has the presentation $\langle \ldots a_{-n}, \ldots, a_1, a_2, \ldots, a_n,\ldots,t \mid a_0^2=1, [a_i,a_j]=1, ta_it^{-1}=a_{i+1}\rangle$. Every finite subpresentation defines a group that has as a quotient one of the following groups $H_n=\langle a_{-n}, \ldots, a_1, a_2, \ldots, a_n,t \mid a_0^2=1, [a_i,a_j]=1, ta_it^{-1}=a_{i+1}\rangle$ for some $n$. The group $H_n$ is an HNN extension of a finite Abelian group $\langle a_{-n},\ldots, a_n\rangle$ with the free letter $t$. Hence $H_n$ is a virtually free group, in particular, $H_n$ contains a non-Abelian free subgroup. Therefore every finite subpresentation defines a group containing a free non-Abelian subgroup, while the Lamplighter group is solvable and thus cannot contain a free non-Abelian subgroup. Similarly lacunary hyperbolic but not hyperbolic groups given by presentations satisfying small cancelation conditions or their generalizations are infinitely presented since every finite subpresentation of their presentation defines a hyperbolic group.
I need to add a very nice characterisation of finitely presented metabelian groups: Robert Bieri and Ralph Strebel, Valuations and finitely presented metabelian groups, In: Proceedings of the London Mathematical Society 3.3 (1980), pp. 439-464, https://doi.org/10.1112/plms/s3-41.3.439