I want to make sure that I am doing this right.
Since $100 = 5\cdot 20$, by Sylow's Third Theorem we have that $n_{5}\equiv 1$ (mod 5) and $n_{5}\mid 20$. So $n_{5}\in\{1,6,11,16,21,26,\dots\}$ and the only possibility is that $n_{5}=1$. Then there is a unique Sylow $5$-subgroup of $G$. Since one of these elements is the identity we have that there are exactly $4$ elements of order $5$ in a group of order $100$.
Best Answer
You didn't use Sylow's theorem correctly though. The decomposition should always be of the form $|G|=p^rm$ where $p$ does not divide $m$. But note that in your example $5$ does divide $20$. But it's easy to fix that, write $100=2^2\times 5^2$. So $n_5\equiv 1$(mod $5$) and $n_5|4$. So it still has to be $n_5=1$ and there is exactly one $5$-Sylow subgroup, and it is of order $25$.
The only groups of order $25$ up to isomorphism are $\mathbb{Z_{25}}$ and $\mathbb{Z_5}\times\mathbb{Z_5}$. The first one contains $4$ elements of order $5$, the second one contains $24$ such elements. Since every element of order $5$ must be contained in a $5$-Sylow subgroup these are all the options for the number of elements of order $5$ in the original group.