c
(c)Kinetic energy\( = \frac{1}{2}m{v^2}\)
As both balls are falling through same height therefore they possess same velocity.
but \(KE \propto m\) (If \(v =\) constant)
\(\frac{{{{\left( {KE} \right)}_1}}}{{{{\left( {KE} \right)}_2}}} = \frac{{{m_1}}}{{{m_2}}} = \frac{2}{4} = \frac{1}{2}\)