Let the reaction be, $\left(\mathrm{P}\right)\to \left(\mathrm{X}\right)+\left(\mathrm{Y}\right)+\left(\mathrm{Z}\right)$

Let mass of each daughter nuclei be $m$, then mass defect can be written as $∆M=M-3m$.

Now the energy released will be in the form of kinetic energy of the daughter nuclei. Therefore, 

$$\begin{gathered} \Delta M{c}^2=\frac{1}{2}m{V}^2+\frac{1}{2}m{V}^2+\frac{1}{2}m{V}^2 \\ \Rightarrow V=\sqrt{\frac{2\Delta M{c}^2}{3m}}=\sqrt{\frac{2\Delta M{c}^2}{M-∆M}}\approx c\sqrt{\frac{2\Delta M}{M}} \end{gathered}$$