A

Step 1: Analyze the forces acting on the mass at the top of the curve. The gravitational force acting downwards is $Mg$, and the normal force $N$ acts upwards. The mass will lose contact with the track when the normal force equals zero: $N = 0$.
Step 2: Set up the equation for centripetal motion at the top of the curve:
$$ \frac{Mv^2}{R} = Mg $$
This simplifies to:
$$ v^2 = gR $$
Step 3: Use the conservation of mechanical energy to find the speed $v$ at height $h$:
$$ mgh = \frac{1}{2} mv^2 + mg(2R) $$
This gives:
$$ h = \frac{1}{2} v^2/g + 2R $$
Substitute $v^2 = gR$ into the equation:
$$ h = \frac{1}{2} (gR)/g + 2R = \frac{1}{2} R + 2R = \frac{5}{2} R $$
Therefore, the minimum value of $h$ for which the mass loses contact with the track is $$ \frac{5}{2} R $$.