= Solution
Let $V_4=\langle u,v\mid u^2=v^2=[u,v]=1\rangle$. Define a surjection
$$
\phi:\pi_1(X_n)\longrightarrow V_4,
\qquad
\phi(a_1)=u,\quad\phi(b)=v,\quad
\phi(a_i)=1\ (i>1).
$$
Its kernel determines a connected regular <four-sheeted cover of a wedge of circles and a real projective plane>, say $\widetilde X_n\to X_n$.
Use the CW structure on $X_n$ with one vertex, one edge for every $a_i$ and for $b$, and one two-cell attached along $b^2$. In the regular cover, the degree-two cellular chain group is
$$
C_2(\widetilde X_n)\cong\mathbb Z[V_4].
$$
The <cellular boundary formula> gives the lifted attaching word $b^2$ the boundary
$$
\partial_2(c)=(1+v)c
$$
in the $b$-edge summand of $C_1$. Therefore
$$
\ker\partial_2
=\mathbb Z(1-v)\oplus\mathbb Z(u-uv)
\cong\mathbb Z^2.
$$
There are no three-cells, so
$$
\boxed{H_2(\widetilde X_n;\mathbb Z)\cong\mathbb Z^2.}
$$
If $X_n$ were homotopy equivalent to a compact connected surface $S$, the subgroup corresponding to this cover would produce a connected four-sheeted cover $\widetilde S$ homotopy equivalent to $\widetilde X_n$. A finite cover of a compact surface is again a compact connected surface. But the <integral top homology of a compact connected surface> has rank at most one, contradicting the displayed rank two. Hence \b[$X_n$ is not homotopy equivalent to a compact connected surface for any $n\geq1$.]
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