In the figure shown, the ball a is released from rest, when the spring is at its natural (un-stretched) length. For the block B of mass M to leave contact with ground at some stage, the minimum mass of A must be :

\(\frac{M}{2}\)

For B just to loose contact : 

\(Mg = kx \Rightarrow x = \frac{Mg}{k}\)

By Work-Energy Theorem :

\(\Rightarrow \frac{-1}{2}kx^2 + mgx = 0\)

$ x = \frac{2mg}{k}$

Put $x = \frac{Mg}{k}$

\(\Rightarrow m = \frac{M}{2}\)