For $n^{3}+5n^{2}$ to be the square of an integer, $\sqrt{n^{3}+5n^{2}}$ must be an integer. We know that $\sqrt{n^{3}+5n^{2}}=\sqrt{n^{2}(n+5)}=\sqrt{n^{2}} \sqrt{n+5}=n \sqrt{n+5}$. For $n \sqrt{n+5}$ to be an integer, $\sqrt{n+5}$ must be an integer. In other words, $n+5$ must be a perfect square. Since $n$ is between 1 and 100, then $n+5$ is between 6 and 105. The perfect squares in this range are $3^{2}=9,4^{2}=16, \ldots, 10^{2}=100$. Thus, there are 8 perfect squares in this range. Therefore, there are 8 values of $n$ for which $\sqrt{n+5}$ is an integer, and thus for which $n^{3}+5n^{2}$ is the square of an integer.