A smooth inclined plane ends in a vertical circular loop, as shown...

Let $v$ be the velocity of the body at the highest point of the circular loop.

At the highest point, the forces acting on the body are its weight $mg$ (downwards) and the normal reaction $N$ (downwards). The net downward force provides the necessary centripetal acceleration:

$N + mg = \dfrac{mv^2}{R}$

Given that the body exerts a force equal to three times its weight on the track at the highest point, we have $N = 3mg$. Substituting this into the equation:

$3mg + mg = \dfrac{mv^2}{R}$

$4mg = \dfrac{mv^2}{R} \Rightarrow v^2 = 4gR$

Applying the principle of conservation of mechanical energy between the point of release and the highest point of the loop (which is at a height of $2R$ from the bottom):

$mgh = mg(2R) + \dfrac{1}{2}mv^2$

Substituting $v^2 = 4gR$ into the energy equation:

$mgh = 2mgR + \dfrac{1}{2}m(4gR)$

$mgh = 2mgR + 2mgR = 4mgR$

Therefore, $h = 4R$.

Comparing this with the given relation $h = \alpha R$, we get $\alpha = 4$.

Answer: $4$