My attempt is :
Since $n_3 > 1$ and $n_3 \equiv 1 \ \ mod \ \ 3 $ and divides 8, then the only possibilty is $n_3 = 4$ and thus $| G:N| = 4$, where $N = N_G(P)$ and $P \in Syl_3(G)$. Then $G/K $ is isomorphic to subgroup of $S_4$, where $K = \cap_{g\in G} N^g$.
We need to show that $K = 1$
Since $|G:N| = 4 $, we see that $|N|$ = 6 and $K \subseteq N$, therefore the possibility of $|K| = 1,2,3,6$
we assume that $K = 2$. Since $|G/K| = 12$ , then exist a normal either 2-sylow subgroup or 3-sylow subgroup. We know that $n_3(G/K) > 1$, however we conclude that $G/K$ has a unique 2- sylow subgroup $S/K$ . Also $|S|$ = 8 Since $|K|$ = 2 and $|S/K|$ = 4 and thus $S$ is a normal Sylow subgroup of $G$, which a contradiction .
I want to help in this arguments
We need to show that $|K|$ is not divisible by 3