As an alternative to your proof (given in a comment, already simplified by Geoff) you could also just use Sylow's theorem and no transfer (surely not intended by Isaacs):
By Sylow the number of $3$-Sylow subgroups has to be either $1$ or $55$. In the first case, you are done, so assume the latter case. Choose two $3$-Sylow subgroups $S\ne T$ such that their intersection $P = S\cap T$ is maximal among intersections of Sylow subgroups.
As $P < \mathrm{N}_S(P)$ and $P < \mathrm{N}_T(P)$, you get by the maximal choice of $P$ that the normalizer $H = \mathrm{N}_G(P)$ does not have a unique normal $3$-Sylow subgroup. As the order of $H$ is $55$ times a power of $3$, $H$ has as many $3$-Sylow subgroups as $G$, each of them properly containing $P$. As $P$ is a maximal intersection of $3$-Sylows of $G$, each $3$-Sylow subgroup of $H$ is contained in a unique $3$-Sylow subgroup of $G$, showing that $P$ is also contained in every $3$-Sylow subgroup of $G$. Hence $P$ is the intersection of all $3$-Sylow subgroups of $G$ and therefore normal in $G$.
Now look at $\bar{G} = G/P$: $\bar{G}$ has $55$ $3$-Sylow subgroups which intersect pairwise trivially (again by maximal choice of $P$) and are self-normalizing (i.e., $\mathrm{N}_{\bar{G}}(\bar{S}) = \bar{S}$). Therefore $\bar{G}$ contains exactly $54$ elements whose order are not a power of $3$.
By Sylow $\bar{G}$ has a unique $11$-Sylow subgroups (there are not enough 3'-elements for more $11$-Sylow subgroups than that), which is normal and centralized by every non-trivial $3$-element as $\mathrm{Aut}(C_{11})$ is cyclic of order $10$, contradicting that the $3$-Sylow subgroups of $\bar{G}$ are self-normalizing.