A finite solvable group $G$ which is not nilpotent and has at most $4$ maximal subgroups satisfies $G/\Phi(G) \cong S_{3},$ where $\Phi(G)$ is the Frattini subgroup, the intersection of all maximal subgroup of $G.$
Suppose $G$ is solvable, not nilpotent, and has at most $4$ maximal subgroups. Suppose also that $\Phi(G) = 1,$ which is no loss of generality. Then $G$ has a maximal subgroup $M$ which is not normal. Then $M$ has at most $4$ conjugates, and there is at least one maximal subgroup of $G$ which is not conjugate to $M.$
Now $M = N_{G}(M)$ by maximality, as $M \not \lhd G.$ We have $[G:M] < 4,$ but we can't have $[G:M]= 2$ as $M$ is not normal. Hence $[G:M] = 3$ and $G/K \cong S_{3},$ where $K$ is the intersection of all $G$-conjugates of $M.$ But then by the isomorphism theorems, there are $4$ maximal subgroups of $G$ containg $K.$ these are the three conjugates of $M,$ together with a normal subgroup $L$ of index $2.$ But this yields $K \leq \Phi(G)$ since $G$ has at most $4$ maximal subgroups. By assumption, $\Phi(G) = 1,$ so that $K = 1$ and $G \cong S_{3}.$