There are at most 4 groups of order 306 = $2 \times 3^2 \times 17$, containing an element of order 9.
I want to prove the above statement. My trial is below;
Let $G$ be such group. If $G$ is abelian, by the fundamental theorem of abelian group, $$G \cong Z_{2} \times Z_{17} \times Z_{3} \times Z_{3} \textrm{ or } Z_{2} \times Z_{17} \times Z_{9}.$$ Thus, $G \cong Z_{2} \times Z_{17} \times Z_{9}$ is one of such group.
Suppose $G$ is nonabelian. Then, by the Sylow's third theorem, $n_3=1$ or $34$, $n_{17}=1$ or $18$, but not both are greater than 1. Thus there are three cases;
Case1 : $n_{3}=1, n_{17}=1$. In this case, we can assume $P_{3} \in Syl_{3}(G)$ is a cyclic group of order 9. Also, $\exists P_{17} \in Syl_{17}(G).$ And $P_3, P_{17}$ are normal subgroup of $G$. Hence $P_3 P_{17}$ is also normal subgroup of $G$. And for any Sylow 2 subgroup, say $P_{2}$, semidirect product of $P_{3}P_{17}$ and $P_{2}$ induces direct product of $P_{3}P_{17}$ and $P_{2}$, thus $G \cong P_{3}P_{17} \times P_{2}$, hence $P_{2}$ should be unique. Thus, this case is one of the nonabelian case containing element of order 9.
Case2 : $n_{3}=1, n_{17}=18$.
Case3 : $n_{3}=34, n_{17}=1$.
I don't know how to proceeds the other two cases. Could you give me some hints?