Let $G$ be a finite group of order $240$. If $G\cong C_4\times A_5$ or $C_2\times C_2\times A_5$, then the all degrees of irreducible $\mathbb{C}$-characters of $G$ are $ [1,1,1,1,~3,3,3,3,3,3,3,3, ~4,4,4,4,~5,5,5,5 ]. $
Conversely, Suppose that $G$ is non-solvable, and the all degrees of irreducible $\mathbb{C}$-characters are $ [1,1,1,1,~3,3,3,3,3,3,3,3, ~4,4,4,4,~5,5,5,5 ]. $
Question: $G\cong C_4\times A_5$ or $C_2\times C_2\times A_5$?
$G$ is non-solvable, so must have $A_{5}$ as a composition factor (as no other non-Abelian simple group has less than $168$). Hence $F(G)$ can have order at most $4$.
If $G$ has no component, then $F(G) = F^{\ast}(G)$ has order $4$, and is centralized by all elements of order $5$, contrary to $C_{G}(F^{\ast}(G)) \leq F^{\ast}(G).$
Hence $G$ has a component (quasi-simple subnormal subgroup). This is isomorphic to either $A_{5}$ or ${\rm SL}(2,5).$ If it is (isomorphic to) ${\rm SL}(2,5),$ it is then normal (it has index two), and ${\rm SL}(2,5)$ has an irreducible character of degree $6$. By Clifford's theorem, $G$ has an irreducible character of degree at least $6,$ which is precluded here.
Hence $G$ has a normal subgroup $L$ isomorphic to $A_{5}.$ Each irreducible character of degree $3$ of $A_{5}$ must extend to an irreducible character of degree $3$ of $G$ (since $G$ has no irreducible character of degree $6$ or more). Hence no element of order $5$ in $L$ can be conjugate to all its non-identity powers. Let $S$ be a Sylow $5$-subgroup of $L$. Then by a Frattini argument, $G = LN_{G}(S) = LC_{G}(S)$ (since $[N_{G}(S):C_{G}(S)] = [N_{L}(S):C_{L}(S)] = 2).$ Now $G$ induces inner automorphisms on $F^{\ast}(G) = L \times O_{2}(G)$ so $G = F^{\ast}(G)$ has one of the listed structures.
The answer is Yes by brute forcing. As there are only 208 groups of order 240, we can check them one by one in GAP:
for p in [1..208]
a:=SmallGroups(240,p);
if not IsSolvableGroup(a) then
c:=CharacterDegrees(a);
if c=[[1,4],[3,8],[4,4],[5,4]] then
Print(StructureDescription(a),"\n");
fi;
fi;
od;
It returns
C4 x A5
C2 x C2 x A5
and the conclusion follows.
EDIT: As the group $G$ has order 240, it must contain $A_5$ as a normal subgroup, so the group $G$ is an extension of $A_5$ by some abelian group with order 4. Concerning the order of the abelian groups with order 4, $G$ can also be described as an extension of some order-120 nonsolvable group by $C_2$. There are only 3 order-120 nonsolvable groups, namely, $SL(2,5)$, $S_5$ and $C_2×A_5$. Both $SL(2,5)$ and $S_5$ have an irreducible representation of degree 6, so the order-120 nonsolvable group in the extension is $C_2×A_5$. Now we just need to check the $C_2$ extensions of $C_2×A_5$, and there are three of them. Two of them are $C_4×A_5$ and $C_2×C_2×A_5$, and the other one has an irreducible representation of degree 6. The classification is now complete.
