$\newcommand\ZZ{\mathbb{Z}}$Using the Hochschild-Lyndon-Serre spectral sequence for the extension $1\to C_3\to S_3\to C_2\to1$ (here $C_n$ is a cyclic group of order $n$) and checking that the action of $C_2$ on $H^3(C_3,\ZZ_3)$ is trivial, we see that restriction induces an isomorphism $\mathrm{res}:H^3(S_3,\ZZ_3)\to H^3(C_3,\ZZ_3),$ and, in particular, by the usual computation of the cohomology of cyclic groups, we see that $H^3(S_3,\ZZ_3)$ is cyclic of order $3$. Moreover, the general theory tells us that the composition $H^3(S_3,\ZZ_3)\xrightarrow{\mathrm{res}} H^3(C_3,\ZZ_3)\xrightarrow{\mathrm{cores}} H^3(S_2,\ZZ_3)$ of the restriction map with the correstriction map is multiplication by the index of $C_3$ in $S_3$, namely $2$: this composition is then an isomorphism too. It follows immediately that to construct a generator of $H^3(S_3,\ZZ_3)$ it is enough to construct a cocycle whose class generates of $H^3(C_3,\ZZ_3)$, which is easy, and then corestrict it to a cocycle on $S_3$, which will generate $H^3(S_3,\ZZ_3)$. To do the latter, there are explicit formulas in textbooks (which I never remember)