( It is given in the question that energy of β-particle emitted is 6.7 MeV.
This is assumed as typo error. It is considered that energy of α-particle emitted is 6.7 MeV )
Energy of daughter nucleus is obtained from conservation of momentum.
If we assume the parent nuclei is initially at rest, then we have, mα × vα + M × V = 0 ......................(1)
where mα is mass of α-particle , vα is the speed of α-particle, M is mass of daughter nucleus and V is speed of daughter nucleus.
If E is the energy of α-particle, then its speed is obtained from, (1/2)mα vα2 = E , then vα = (2E/mα )1/2
Hence Speed of daughter nucleus , V = - ( mα × vα ) / M = - [ mα × (2E/mα )1/2 ] / M
Hence recoil energy of daughter nucleus = (1/2) M V2 = ( mα × E ) / M = ( 4 × 6.7 )/214 = 0.125 MeV