smc

An inverse variation can be expressed in the form $xy=n$, where $n$ is any number (except perhaps zero). Since $5y$ varies inversely with the square of $x$, this particular one will be $5y{x}^2=n$. We can plug in $16$ for $y$ and $1$ for $x$, which makes $n=80$. The equation is now $5y{x}^2=80$. When $x=8$, we can solve for $y$ using the equation $320y=80$, which makes $y$ $\boxed{\frac{1}{4}}$.