Explanation

(c) For minimum mass of $m$, mass $M$ breaks off contact when elongation in spring is maximum
At the time of break off, block $A$ is at lowest position and its speed is zero. At an instant $t_{1}$
$mg - kx = ma$
$v\frac{dv}{dx} = \frac{mg - kx}{m}$
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$\int_{0}^{0}{vdv} = \int_{0}^{x}{\left( g - \frac{k}{m}x \right)dx}$
Where $x_{0}$ is maximum elongation is spring
$0 = g\ x_{0} - \frac{kx_{0}^{2}}{2m}$
$x = \frac{2mg}{k}$
At the time of break off of block $B$
$Mg = kx_{0}$
$Mg = 2mg$
$m = \frac{M}{2}$