Solution: We need to determine the number of subgroups of order $5$. Note that any subgroup of order $5$ is cyclic, so the order of each element other than identity is $5$.

Also note that if $g$ and $h$ are two elements of order $5$, then we have either $\left\langle g \right\rangle =\left\langle h \right\rangle $ or $\left\langle g \right\rangle \cap \left\langle h \right\rangle =\emptyset$. This follows from the fact that the intersection $\langle g \rangle \cap \langle h \rangle$ is a subgroup of the group $\langle g \rangle $, and thus the order of the intersection is either $1$ or $5$.

It is given that there are $28$ elements in $G$ with order $5$. Since each subgroup has $4$ number of elements with order $5$, the total number of subgroups of order $5$ will be $\frac{28}{4}=7$.