Since ball is at a specific height it possess potential energy
= mgh
where m is mass, g is gravity and h is height
Initial energy of ball = mg $ h_1 $
= 20 mg
Final energy of ball = $ mgh_2 $
=10 mg.
Therefore, the loss in energy
$ \frac{initial\, energy - final\, energy}{initial energy} \times 100 $
= $ \frac{20mg - 10 mg}{20 mg}\times 100 $
=50%