I don't really think that this is a research level question, but here is quick answer.
Since the Schur Multiplier of the perfect group ${\rm PSL}(2,q)$ has order $2$, $G$ must have a normal subgroup $N$ of index $2$ isomorphic to $C_3 \times {\rm PSL}(2,q)$. Since ${\rm PGL}(2,q) \setminus {\rm PSL}(2,q)$ contains an element of order $2$, we have $G = \langle N,t \rangle$, with $t^2=1$. The direct factors of $N$ are both characteristic in $N$ and hence normalized by $t$, and clearly $\langle t, {\rm PSL}(2,q) \rangle = {\rm PGL}(2,q)$, so the group $G$ is detemrined by the conjugation action of $t$ on $H=C_3$.
There are two possibilities for that, $t$ can either centralize of invert the elements of $H$, giving two isomorphism classes of groups.