$G\cong C_4\times A_5$ or $C_2\times C_2\times A_5$? 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 $C_2\times C_2\times A_5$?

 A: 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.
A: $G$ 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.
