A nucleus of radium-226 at rest emits an alpha particle and becomes radon-222…

Step 1. The lost mass. $226.025410\ \text{u} - 222.017578\ \text{u} - 4.002603\ \text{u} = 0.005229$ u.

Step 2. The energy. $Q = 0.005229\ \text{u} \times 931.494\ \text{MeV/u} = 4.871$ MeV.

Step 3. The sharing. Momentum is conserved from rest, so $m_\alpha v_\alpha = m_{Rn}v_{Rn}$. Kinetic energy is $K = \dfrac{p^2}{2m}$, with the same $p$ for both. So $\dfrac{K_\alpha}{K_{Rn}} = \dfrac{m_{Rn}}{m_\alpha}$, and $K_\alpha = Q\,\dfrac{m_{Rn}}{m_{Rn} + m_\alpha} = 4.871\ \text{MeV} \times \dfrac{222.017578}{226.020181} = 4.785$ MeV.

Step 4. The radon nucleus. $K_{Rn} = 4.871\ \text{MeV} - 4.785\ \text{MeV} = 0.086$ MeV.