Answer
$ h=2.7(R-r)$
Work Step by Step
First of all, we need to draw the force diagram of the ball at its highest point $h$ and when it is at the top of the loop-the-loop track, as we see below.
To find the minimum height, we need to find the minimum velocity that makes the ball barely complete the circle.
To find the minimum velocity of the marble that barely make it complete the vertical circle. This occurs when the marble is barely touching the track at the top point on the track, as we see in the figure below, which means that the normal force at this point is zero.
Applying Newton's second law at the top point of the circle;
$$\sum F_3=mg+\overbrace{F_{n3}}^{\approx \;0}=ma_r$$
where $a_r$ is the radial acceleration which is given by $v^2/r_c$ where $r_c$ here is given by $R-r$
$$ \color{red}{\bf\not}mg =\dfrac{ \color{red}{\bf\not}mv_3^2}{R-r}$$
Hence,
$$v_3^2=g(R-r) $$
Recalling that $v=\omega r$ and in our case the marble is also rotating around its center of mass, So $v_3=\omega_3 r$
$$(\omega_3 r)^2=g(R-r) $$
$$\omega_3^2 r^2=g(R-r) $$
Hence,
$$\omega_3^2 =\dfrac{g(R-r)}{r^2}\tag 1 $$
Now we need to find the height that gave us this speed by applying the conservation of energy principle.
$$E_{i}=E_f$$
where $E_i$ is the energy of the marble at the top of the hill and $E_f$ is its energy at the top of the circle (at position 3 in the figure below).
$$K_i+U_{ig}=K_{f}+U_{gf}$$
We know that the marble start rolling down, without friction, from rest.
This means that the initial kinetic energy is zero.
Noting that $y_i=h+r$ since we are measuring the height relative to the meable\s center of mass; and hence, $y_f=2R-r$
$$0+mg(h+r)=\frac{1}{2}mv_3^2+\frac{1}{2}I\omega_3^2+mg(2R-r)$$
We know that the moment of inertia of the marble (solid sphere) is given by $\frac{2}{5}mr^2$
$$mg(h+r)=\frac{1}{2}mv_3^2+\frac{1}{2}\left(\frac{2}{5}mr^2\right) \omega_3^2+mg(2R-r)$$
As we mentioned above, $v_3=\omega_3 r$;
$$ \color{red}{\bf\not}mg(h+r)=\frac{1}{2} \color{red}{\bf\not}m(\omega_3 r)^2+ \frac{1}{5} \color{red}{\bf\not}mr^2 \omega_3^2+ \color{red}{\bf\not}mg(2R-r)$$
$$ g(h+r)=\frac{1}{2} \omega_3^2 r^2+ \frac{1}{5} r^2 \omega_3^2+ g(2R-r)$$
$$ gh+gr = \frac{7}{10} \omega_3^2r^2+ g(2R-r)$$
Plugging from (1);
$$ gh+gr = 0.7 \left[\dfrac{g(R-r)}{ \color{red}{\bf\not}r^2}
\right] \color{red}{\bf\not}r^2+ g(2R-r)$$
$$ \color{red}{\bf\not}gh+ \color{red}{\bf\not}gr=0.7 \color{red}{\bf\not}gR-0.7 \color{red}{\bf\not}gr+2 \color{red}{\bf\not}gR- \color{red}{\bf\not}gr$$
$$ h =0.7 R-0.7 r+2 R-2 r=2.7R-2.7r$$
Therefore,
$$\boxed{h=2.7(R-r)}$$
