This is an old comp question
Here's what I got: We can use the Classification of Compact Surfaces to find the Fundamental Group.
This is a non-orientable surface, with the Mobius band going from the edges identified by $b$. There is 1 face, and 4 edges: Edges given by identifying $a$ and $b$, as well as the two unidentified edges. I computed the number of vertices to be 2. All the vertices apart from the one connecting the two unidentified edges should be sent to the same vertex in the quotient. So we should get
$$\chi(Y) = V-E+F = 2-4+1 = -1$$
Via the Classification of Compact Surfaces we know that for a non-orientable surface the genus is given by $$\chi(Y) = 2-g \implies g=3$$ So $$Y\cong \mathbb{R}P^2 \# \mathbb{R}P^2 \# \mathbb{R}P^2$$
Thus $$\pi_1(Y) \cong \langle a,b,c : a^2b^2c^2 \rangle$$
Does this look correct? I often have trouble computing the number of vertices for these classification problems, so I worry that I've made a mistake. Thank you in advance for your help.