If you are adding 13 numbers and the average is $4.6$, this is equivalent to their sum being $4.6\times 13$. Unfortunately, that gives $59.8$, which is impossible to achieve if your original numbers were all integers. Which suggests (to me, at least) that you are in fact approximating the average, and that the actual average you want is $\frac{60}{13} = 4.615384615\ldots$
So, you want to know how many ways you can obtain $60$ as a sum of exactly $13$ integers, all taken from ${1,2,3,4,5}$. This is equivalent to counting the number of $5$-partitions of $60$ of length exactly $13$.
It can probably be done with generating functions, but here you can do it from first principles.
Note that the maximum sum you can possibly get with $13$ numbers taken from ${1,2,3,4,5}$ is $65$. So the partitions you want are in 1-to-1 correspondence with the partitions of $5$ other than the trivial one ($5=5$).
For instance, the partition $5=1+4$ gives the partition $60 = 11(5) + 4 + 1$ (subtract $1$ from one summand, $4$ from another summand). The partition $5=2+3$ yields the partition $60=11(5) + 3 + 2$. Etc. So you have: \begin{align*} 5 &= 1+4 &\qquad 60 &= 11(5) + 4 + 1;\\ 5 &= 2+3 &\qquad 60 &= 11(5) + 3 + 2;\\ 5 &= 1+1+3 &\qquad 60 &= 10(5) + 4 + 4 + 2;\\ 5 &= 1+2+2 &\qquad 60 &= 10(5) + 4 + 3 + 3;\\ 5 &= 1+1+1+2 &\qquad 60&= 9(5) + 4 + 4 + 4 + 3;\\ 5 &= 1+1+1+1+1 &\qquad 60&= 8(5) + 4 + 4 +4 + 4 + 4; \end{align*} where $60 = k(5) +\cdots$ means that you take the number $5$ $k$ times, and then the other numbers.
So there are exactly $6$ ways of doing it.