= Solution
A <covering space> is a map $p:\widetilde X\to X$ such that every $x\in X$ has an open neighbourhood $U$ for which
$$
p^{-1}(U)=\bigsqcup_{\lambda}U_\lambda,
$$
with every restriction $p:U_\lambda\to U$ a homeomorphism.
Let $N=\langle\!\langle a\rangle\!\rangle$ in
$\pi_1(Y)=\langle a,b\mid a^2b^{-3}\rangle$. Then
$$
\pi_1(Y)/N\cong\langle b\mid b^3=1\rangle\cong C_3.
$$
The subgroup is normal of index three, so the <degree of a connected covering> shows that every fibre $p^{-1}(y)$ has exactly three points.
The <cell complex of a covering from a coset graph> gives an explicit model. Take vertices $v_0,v_1,v_2$ with indices modulo three; put an $a$-loop $A_i$ at every $v_i$, and a directed $b$-edge $B_i:v_i\to v_{i+1}$. Attach a 2-cell at each $v_i$ along
$$
A_i^2B_{i-1}^{-1}B_{i-2}^{-1}B_i^{-1}.
$$
Map every $v_i$ to $y_0$, every $A_i$ to the $a$-cell, every $B_i$ to the $b$-cell, and each 2-cell homeomorphically to the unique 2-cell of $Y$. Its attaching word is the lift of $a^2b^{-3}$ beginning at $v_i$. This defines the required connected three-sheeted covering.
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