Answer
$ 8.7\times 10^{-34}$
Work Step by Step
First, we need to find the moment of inertia of the uniform rod (baton) with the end masses. Assuming that the length of the rod is $R$, as you see in the figure below.
The net moment of inertia around the center of mass of the system is given by
$$I_{net}=\frac{1}{12}m_{rod}R^2+m_{mass}\left(\dfrac{R}{2}\right)^2+m_{mass}\left(\dfrac{R}{2}\right)^2$$
Thus,
$$I_{net}=\frac{m_{rod}R^2}{12}+ \dfrac{ m_{mass}R^2}{2} \tag 1$$
We know that the kinetic energy of the baton is given by
$$KE=\frac{1}{2}I\omega^2\tag 2$$
We also know that the quantized rotational energy levels are given by
$$E=l(l+1)\dfrac{\hslash^2}{2I}$$
and since we assume that the quantum number will be very large,
$$E=l(l+1)\dfrac{\hslash^2}{2I} \approx l^2\dfrac{\hslash^2}{2I} $$
Solving for $l$;
$$l=\sqrt{\dfrac{2IE}{\hslash^2}}=\sqrt{\dfrac{8\pi^2 IE}{h^2}}$$
Plugging from (2);
$$l= \sqrt{\dfrac{8\pi^2 I\;\frac{1}{2}I\omega^2 }{h^2}}$$
$$l= \sqrt{\dfrac{4\pi^2 I^2 \omega^2 }{h^2}}$$
$$l= \dfrac{2\pi I \omega }{h } $$
Plugging from (1) and recall that $\omega=2\pi f$
$$l= \dfrac{4\pi^2f \left[\dfrac{m_{rod}R^2}{12}+ \dfrac{ m_{mass}R^2}{2}\right] }{h } $$
Plugging the known;
$$l= \dfrac{4\pi^2(1.8) \left[\dfrac{ (0.230)(0.32)^2}{12}+ \dfrac{(0.380)(0.32)^2}{2}\right] }{6.626\times 10^{-34}} $$
$$l=\bf 2.3\times10^{33}$$
So, the differences between rotational energy levels for a spinning baton compared to the energy of the baton is given by
$$\dfrac{\Delta E}{E}=\dfrac{\dfrac{\hslash^2 l}{I}}{ \dfrac{l^2\hslash^2}{2I}}$$
$$\dfrac{\Delta E}{E}= \dfrac{\color{red}{\bf\not}\hslash^2 l}{\color{red}{\bf\not}I}\cdot \dfrac{2\color{red}{\bf\not}I}{l^2\color{red}{\bf\not}\hslash^2} =\dfrac{2}{l}$$
Plugging from above;
$$\dfrac{\Delta E}{E} =\dfrac{2}{2.3\times10^{33}}=\color{red}{\bf 8.7\times 10^{-34}}$$
Thus,
$$ \Delta E =\dfrac{2}{2.3\times10^{33}}=\color{red}{\bf 8.7\times 10^{-34}} E$$
This is a very small difference which will be not detectable.
