$$\begin{gathered} restoring force of spring is F=K{x}_0 \\ in equilibrium, K{x}_0=mg \\ {x}_0=\frac{mg}{K}\text{---(1)} \\ apply law of conservation of linear momentum to perfectly inelastic collision \\ {(p_1+p_2)}_{before}={(p_1+p_2)}_{after} \\ 0+mv=(m+m)V_{final} \\ {V}_{final}=\frac{v}{2}---\left(2\right) \\ K{E}_{blocks}+P{E}_{spring}=work done \\ \hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\frac{1}{2}\times 2m x {V^2}_{final}+\frac{1}{2}K{x_0}^2=F.x_0 \\ \therefore \frac{1}{2}\times 2m{\left(\frac{V}{2}\right)}^2+\frac{1}{2}K{\left(\frac{mg}{K}\right)}^2=2mg\left(\frac{mg}{K}\right) \end{gathered}$$      
  
          $\Rightarrow V=\sqrt{\frac{6m{g}^2}{K}}$