As ball is dropped from 12 m,

$Initial\ velocity\ before\ impact\ = \sqrt{2gh}=\sqrt{2g\times12} \ \ m/s$

As it rebounds to a height of 4m,

$Velocity\ after\ impact\ = \sqrt{2gh}=\sqrt{2g\times4} \ \ m/s$

$\displaystyle \therefore{}e=\dfrac{\sqrt{2g\times{}4}}{\sqrt{2g\times{}12}}=\sqrt{\dfrac{4}{12}}=0.577$