= Solution
A closed connected <three-manifold> has a <handle decomposition> with one zero-handle and one three-handle. To arrange this, cancel the zero-handles along a spanning tree of connecting one-handles; perform the corresponding construction in the dual <handle decomposition> to consolidate the three-handles. This does not require the <three-manifold> to be orientable.
If there are $r$ one-handles and $s$ two-handles, the <Euler characteristic> is
$$
\chi(Y)=1-r+s-1=s-r.
$$
<Mod-two Poincare duality> pairs the <Betti numbers> of a closed odd-dimensional <manifold>, giving $\chi(Y)=0$. Thus $r=s$. The one-handles give $r$ <generators of a group> of the <fundamental group>, the attaching circles of the two-handles give $r$ <relators>, and the three-handle does not change the <fundamental group>. Therefore \b[$\pi_1(Y)$ admits a balanced presentation].
Now let
$$
G=\langle x_1,\ldots,x_r\mid R_1,\ldots,R_s\rangle
$$
be any finite <group presentation>. Start with a four-dimensional zero-handle and $r$ one-handles. Its <fundamental group> is the <free group> on the $x_i$. Represent the finitely many <relators> by disjoint embedded circles in its three-dimensional boundary; a small perturbation makes the circles disjoint. Choose <framings of an embedded sphere> and attach $s$ two-handles. The resulting compact connected oriented four-manifold $W$ has $\pi_1(W)\cong G$ by the <Seifert-van Kampen theorem>.
The map $\pi_1(\partial W)\to\pi_1(W)$ is surjective. One way to see this is to turn the <handle decomposition> upside down: the relative handles based on $\partial W$ have indices $2,3,4$, and so add no <fundamental group> generators. Also $\partial W$ is connected, since surgery on embedded circles in a connected three-manifold preserves connectedness.
Take the <double of a manifold>
$$
M=W\cup_{\partial W}(-W).
$$
It is a closed connected smooth oriented four-manifold. The <Seifert-van Kampen theorem> gives
$$
\pi_1(M)\cong G*_{\pi_1(\partial W)}G.
$$
Both maps from $\pi_1(\partial W)$ are the same surjection, under the natural identification of the two copies. The pushout identifies the two copies of every element of $G$ and imposes no new relations: the fold homomorphism is inverse to either inclusion. Thus
$$
\boxed{\pi_1(M)\cong G.}
$$
This is <four-manifold realization of finitely presented groups>.
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