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.

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.

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 four 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.

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.