Elastic potential energy is given by formula
[tex]U = \frac{1}{2} kx^2[/tex]
here we know that
[tex]k = 800 N/m[/tex]
[tex]x = 6 m[/tex]
Now using above formula we have
[tex]U = \frac{1}{2}(800)(6^2)[/tex]
[tex]U = 14400 J[/tex]
So elastic potential energy in the chord is 14400 J