$\DeclareMathOperator{\Ima}{Im}$ By the first isomorphism theorem of groups, we have that
$$G/\ker g \cong g(G) \leq S_4 $$
Since the group $S_4$ is solvable, we know that every subroup of $S_4$ is solvable. Hence $g(G)$ is solvable, and so is $G/\ker g$.
So, in order to show that $G$ is solvable, it suffices to show that $\ker g$ is solvable.
We have $|\ker g|=24=2^33$. So:
$$n_3 \equiv 1 \pmod 3 \ \ \ \ \ \ \text{and} \ \ \ \ n_3 \ | \ 8,$$
$$n_2 \equiv 1 \pmod 2 \ \ \ \ \ \ \text{and} \ \ \ \ n_2 \ | \ 3,$$
where $n_3$ represents the number of Sylow $3$-subgroups of $\ker g$, and $n_2$ represents the number of Sylow $2$-subgroups of $\ker g$. Hence $n_3 = 1$ or $n_3=4$ and $n_2 = 1$ or $n_2=3$.
If $n_3 = 1$, then there is a unique Sylow $3$-subgroup, $P_3$, which is a normal subroup of $\ker g$.
So we have the subnormal series
$$\{1\} \trianglelefteq P_3 \trianglelefteq \ker g $$
from which we can conclude that the group $\ker g$ is solvable, since $|\ker g / P_3|=8=2^3$ is a prime power, hence the group $\ker g / P_3$ is abelian.
Similarly, one could prove that $\ker g$ is solvable if $n_2 = 1$.
The problem is, I don't know how to exclude the case $n_3 = 4$ and $n_2 = 3$.
Any help would be greatly appreciated.
Best Answer
Let $N=\mathrm{ker}(G)$. Assume $n_2=3$. Then $N$ acts by conjugation on the set of $2$-Sylow subgroups, yielding a homomorphism $N\to S_3$. Because the action is transitive, the image cannot be trivial; the image must be either $A_3$ (cyclic of order $3$) or all of $S_3$. In both cases we have a solvable image, so again we can just go down to the kernel. If the image is cyclic of order $3$, then the kernel is of order $8$, and being a $2$-group, it is solvable (nontrivial center). If the image has order $6$, then the kernel has order $4$ and so must be abelian, hence solvable.