\(mv = (2m + M) v'\)
\(v' = \frac{{mv}}{{2m + M}}\)
Initial energy
\(\frac{1}{2}\) \(m{v^2}\)
Final energy
\(\frac{1}{2}\,\left( {2m + M} \right){\left( {\frac{{mv}}{{2m + M}}} \right)^2}\)
\(Initial\, kinetic\, energy - Final\, Kinetic\, energy =\) \(\frac{5}{6}\) of initial kinetic energy.
After solving, we get, \(\frac{M}{m} = 4.\)