= Solution
Fix punctures $p,q,r$. There is one isotopy class $\ell_p$ of essential arc from $p$ back to $p$: by the <Jordan curve theorem>, it separates $q$ from $r$, and any two such arcs are isotopic relative to the punctures.
Altogether $\mathcal A(S_{0,3})$ has six vertices:
$$
a_{pq},a_{pr},a_{qr},\ell_p,\ell_q,\ell_r.
$$
The joining arcs $a_{pq},a_{pr},a_{qr}$ span a triangle. For each puncture $p$, the three vertices
$$
\ell_p,\ a_{pq},\ a_{pr}
$$
span another triangle. These are all the maximal simplices: $\ell_p$ must cross $a_{qr}$, and returning arcs based at distinct punctures cannot be made disjoint.
Every orientation-preserving permutation of the three punctures is realizable, while a homeomorphism acting trivially on the punctures is isotopic to the identity. Thus
$$
\operatorname{Mod}(S_{0,3})\cong S_3.
$$
It permutes the labels in the displayed description. There are two vertex orbits, the three joining arcs and the three returning arcs, as summarized by the <arc complex of the three-punctured sphere>.
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