I have a task which asks me to show, that every group of order 300 is solvable. But this should be false. When you take the direct product of the alternating group $A_5$ with $\mathbb{Z}/5\mathbb{Z}$, hence $A_5\times\mathbb{Z}/5\mathbb{Z}$, this group should not be solvable, since $A_5$ is not solvable.
So there should be at least one group of order 300, which is not solvable.
You are right, $A_5\times C_5$ is not solvable, because $A_5$ is not. This is easily seen if you know that $G$ is soluble if and only if its composition factors are cyclic, since $A_5$ is non-cyclic composition factor of $A_5\times C_5$. If you aren't allowed to assume this fact then consider any proper normal subgroup $N$ of $A_5\times C_5$. $N\cap A_5$ is also normal so is either $A_5$ or trivial, so either $N=A_5$ or $G/N\cong A_5$ so any subnormal series of $A_5\times C_5$ contains $A_5$ which is non-abelian.
To take this a bit further, $A_5\times C_5$ is indeed the only non-solvable group of order $300$. To see this you could just check the list in GAP or MAGMA, which I assume was done for the comments, but lets have fun and prove it by hand! Assume $G$ is such a group and note that $G$ must have a simple composition factor of order dividing $300$. Checking this list the only possibility is $A_5$.
If $A_5$ is not a subgroup of $G$ then it must be a quotient with $C_5\cong N\trianglelefteq G$ so $G/N\cong A_5$. $C_5$ is abelian, so conjugation in $G$ defines an action of $A_5$ on $C_5$. Since $1\in C_5$ is fixed under this action $A_5$ acts on the non-trivial elements of $C_5$, but there is no non-trivial action of $A_5$ on a set of $4$ elements so $G$ is a central extension of $A_5$. The Schur Multiplier of $A_5$ is $C_2$ which implies that $G$ must be isomorphic to $A_5\times C_5$.
If $A_5$ is a subgroup of $G$ then conjugation defines a map $G\to{\rm Aut}(A_5)\cong S_5$, the image of which must be $A_5$ so for any $g\in G$ there is some $h\in A_5$ with $x^g=x^h$ for all $x\in A_5$ (where $x^g=g^{-1}xg$). Fix $g\notin A_5$ so $gh^{-1}\notin A_5$. We have $x^{gh^{-1}}=1$ for all $x\in A_5$ and $(gh^{-1})^5\in A_5$ so we must have $\langle gh^{-1}\rangle\cong C_5$ and $gh^{-1}\in Z(G)$. Hence $G\cong A_5\times C_5$.