Coverings of connected sum of four copies of $\mathbb{R}P^2$ G.Baumslag in one of his papers asserts that a group $G = \langle a,b,c,d | a^2b^2c^2d^2 = 1 \rangle$ contains all fundamental groups of closed compact orientable surfaces of genus $g\geq 2$? I think it can be proved using covering theory. The group $G$ is the fundamental group of $\mathbb{R}P^2\sharp\mathbb{R}P^2\sharp\mathbb{R}P^2\sharp\mathbb{R}P^2$. We know that $M_k$ (compact orientable surface of genus $k$) covers $M_2$ by $\mathbb{Z}/{(k-1)}$-action. So the sufficient condition would be that $M_2$ covers $\mathbb{R}P^2\sharp\mathbb{R}P^2\sharp\mathbb{R}P^2\sharp\mathbb{R}P^2$?
However, I know that $M_k$ double covers the connected sum of $k+1$  projective planes $\sharp^{k+1} \mathbb{R}P^2$. In particular, $M_3$ (not $M_2$) covers $\mathbb{R}P^2\sharp\mathbb{R}P^2\sharp\mathbb{R}P^2\sharp\mathbb{R}P^2$. It follows that $G$ contains groups of all orientable surfaces of genus $2k+1$ with $k\geq 1$. What to do with even genus?

$M_2$ cannot cover your space, since they both have the same (non-zero) Euler characteristic.
EDIT: Just to expound a bit, this means $\pi_1(M_2)$ is not a subgroup of your group.
DOUBLE EDIT: This also rules out all even-genus closed orientable surfaces, since they would have to go through the orientable double-cover of your space, which has Euler characteristic $-4$.
