Given, $\left(p^2+q^2\right)x^2-2q\left(p+r\right)x+q^2+r^2=0$

On simplifying we get, ${\left(px-q\right)}^2+{\left(qx-r\right)}^2=0$

$\Rightarrow px-q=0 \& qx-r=0$

$\Rightarrow x=\frac{q}{p}=\frac{r}{q}$

$\Rightarrow x=\frac{q}{p}=\frac{r}{q}=4$    [because roots of equation $x^2-2x-8=0$ are $4$ or $-2$]

As $p,q,r$ are positive, so $x$ must be $4$.

Now, $q=4p$ and $r=4q=16p$

So, $\frac{q^2+r^2}{p^2}=\frac{{\left(4p\right)}^2+{\left(4\times 4p\right)}^2}{p^2}=16+256=272$.