My attempt at a solution:
Because of the Sylow theorems, $G$ has either one or three Sylow $2$-subgroups. If it has one, then $G$ has just one Sylow $2$-subgroup (which has order $8$), so it is normal, and we are done.
Now suppose that $G$ has three Sylow $2$-subgroups. I tried to show that the center of the group, $Z(G)$, has order $4$ or $12$ if $G$ not Abelian. In this case the result follows again because of the existence of a Sylow $2$-subgroup in $Z(G)$.