Answer
The magnitude of the horizontal force at point $Q$ is $~~1.96\times 10^{-2}~N$
Work Step by Step
We can use conservation of energy to find an expression for the total kinetic energy at point $Q$ which is at a height of $R$:
$U_2+K_2 = U_1+K_1$
$K_2 = U_1+0-U_2$
$K_2 = Mg(6.00~R)-MgR$
$K_2 = 5.00~MgR$
We can write an expression for the rotational inertia of a solid sphere:
$I = \frac{2}{5}Mr^2$
We can find an expression for the total kinetic energy of the ball when it has a speed $v$:
$K = \frac{1}{2}Mv^2+\frac{1}{2}I\omega^2$
$K = \frac{1}{2}Mv^2+\frac{1}{2}(\frac{2}{5}Mr^2)(\frac{v}{r})^2$
$K = \frac{1}{2}Mv^2+\frac{1}{5}Mv^2$
$K = \frac{7}{10}Mv^2$
We can find an expression for $v^2$ at point $Q$:
$K = \frac{7}{10}Mv^2 = 5.00~MgR$
$v^2 = \frac{50.0~gR}{7}$
At point $Q$, the ball is moving around a loop. The track exerts a centripetal force on the ball to keep the ball moving around in a circle. The direction of a centripetal force is towards the center of the circle.
We can find the magnitude of this horizontal force:
$F = \frac{Mv^2}{R}$
$F = \frac{M(\frac{50.0~gR}{7})}{R}$
$F = \frac{50.0~Mg}{7}$
$F = \frac{(50.0)~(0.280\times 10^{-3}~kg)(9.8~m/s^2)}{7}$
$F = 1.96\times 10^{-2}~N$
The magnitude of the horizontal force at point $Q$ is $~~1.96\times 10^{-2}~N$
