A group $G$ is called a 2-Frobenius group if $G=ABC$, where $A$ and $AB$ are normal subgroups of $G$, $AB$ is a Frobenius group with kernel $A$ and complement $B$ and $BC$ is a Frobenius group with kernel $B$ and complement $C$.
Let $A$ be a nilpotent group of order $2^4.3^4=1296$ and order $B$ equal to $5$ and order $C$ equal to $4$. So we know that $BC$ is the only Frobenius group of order $20$. Now, I would like to know that whether we can find all such 2-Frobenius group $G=ABC$ of order 25920? Actually I am interested in finding $Irr(G)$ as well. (Irr(G) is the set of irreducible characters of group $G$).
The smallest dimensional modules for a cyclic group of order $5$ over both ${\mathbb F}_2$ and ${\mathbb F}_3$ have dimension $4$, so $A$ must be a direct product of elementary abelian groups of orders $2^4$ and $3^4$. Also, $BC$ has a single faithful $4$-dimensional module over both ${\mathbb F}_2$ and ${\mathbb F}_3$, so there is a unique $2$-Frobenius group $ABC$ of the type you describe.
You get $2$-Frobenius groups $A_1BC$ and $A_2BC$ as the semidirect product of these two irreducible modules with $BC$, where $|A_1|=2^4$, $|A_2|=3^4$, and $ABC$ is a subdirect product $(A_1 \times A_2)BC$ of $A_1BC$ and $A_2BC$. Hence $ABC$ has an intransitive permutation representation of degree $97$.
You should probably use GAP (or Magma) to compute the character table of $ABC$. It has $111$ conjugacy classes.