Which group of order 24 is this group?

Notice that $G$ is generated by $m^4$, $m^3$ and $g$.

Also $gm^4g^{-1}=m^8$, so $\langle m^4,g\rangle\cong S_3$ (e.g. identify $m^4$ with $(1,2,3)$ and $g$ with $(2,3)$ ).

For $m^3$, $gm^3g^{-1}=m^3$, so $m^3\in Z(G)$.

Hence $G\cong S_3\times C_4$.

Here is a proof using Sylow theory.

$C_2$ acts trivially on the copy of $C_4$ in $C_{12}$ so the Sylow $2$-subgroup of $G$ is isomorphic to $C_2\times C_4$. Also, $G$ has only one Sylow $3$-subgroup (a Sylow $3$-subgroup of $G$ is characteristic in $C_{12}$ which is normal in $G$). Then $G\cong C_3\rtimes(C_2\times C_4)$ where $C_4$ acts trivially on $C_3$ and where $C_2$ acts nontrivially on $C_3$. This is isomorphic to $(C_3\rtimes C_2)\times C_4$ where $C_2$ acts nontrivially on $C_3$. Thus, $G\cong S_3\times C_4$.