May the Potential Energy be with you!

Nice one.
Consider everthing in SI units in my solution.

For a quarter circular ring, center of mass lies at $\displaystyle\bigg(\frac{2R}{\pi}, \frac{2R}{\pi}\bigg)$ .

Therefore, potential energy

$\phi = mgy$ $\phi = mg\frac{2R}{\pi}$ $\phi = \frac{2mgR}{\pi}$ $\phi = \frac{2(5)(10)(\pi)}{\pi}$ $\phi = 100J$

Let us solve the problem in general.

Let the mass of the chain be $m$ , radius of the quarter circle be $r$ , acceleration due to gravity be $g$ .

Consider a small element of the chain of length $dl=rdα$ at a point at an angular position $α$ from the base line, which is the surface of the earth. Height of this point above the earth's surface is $h=r\sin α$ .

Length of the chain is $l=\dfrac {πr}{2}$ , so that the mass per unit length is $\dfrac {2m}{πr}$ .

Mass of the element is $dm=\dfrac ml \times rdα=\dfrac {2m}{π}dα$ ,

and it's potential energy is

$d(P.E.)=\dfrac {2mgr}{π}\sin αdα$

So the potential energy of the entire chain is

$P. E. =\displaystyle \int_0^{\frac π2} \dfrac {2mgr}{π}\sin αdα=\dfrac {2mgr}{π}$

Now for the given problem, $m=5,g=10,r=π$

Hence the required potential energy is

$P. E. =\dfrac {2\times 5\times 10}{π}=\boxed {100}$ Joules.