Equipartion Theorem rotation question

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I have absolutely NO IDEA what d is asking or how to do e. A and B are simple, i used the formula E = hbar^2(l(l+1) / 2I (I is moment of inertia) to get c. Can anyone help me with D and E please?

Consider the model of a diatomic gas fluorine (F2) shown in Figure 9.3.

(Figure is 2 atoms connected by an imaginary "rod" along the z axis)
Figure 9.3

(a) Assuming the atoms are point particles separated by a distance of 0.14 nm, find the rotational inertia Ix for rotation about the x axis.
3.1e-46 kg·m2
(b) Now compute the rotational inertia of the molecule about the z axis, assuming almost all of the mass of each atom is in the nucleus, a nearly uniform solid sphere of radius 3.2 x 10^-15 m.
2.58e-55 kg·m2
(c) Compute the rotational energy associated with the first (l = 1) quantum level for a rotation about the x axis.
3.6e-23 J
(d) Using the energy you computed in (c), find the quantum number script i needed to reach that energy level with a rotation about the z axis.

(e) Comment on the result in light of what the equipartition theorem predicts for diatomic molecules.
 
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I just had this question on my homework. Part (d) is bullsh*t, simply. It doesn't actually mean an integer when it says 'quantum number'.

For me, I had 8.9e-25 for part (c), and 3.188e-54 for (b), so part (d) looked like:

[itex]E_{rot} = \frac{\hbar^2 l (l + 1)}{2I}[/itex]
Rearrange...
[itex]\frac{2 E_{rot} I}{\hbar^2} = l (l + 1)[/itex]
Plug in...
[itex]\frac{2 (8.9e-25) (3.188e-54)}{(1.05e-34)^2} = l (l + 1)[/itex]

Solving for l gives 5.147e-10 which I cannot understand as a quantum number, since l = 0, 1, 2, ...n-1.

Conceptually, the l = 1 energy level for rotation about the X axis has to be a much lower energy level than the l = 1 energy level for rotation about the Z axis since the moment of inertia for the Z axis is so much smaller.

Bogus question, or perhaps I missed something. Anyway, WebAssign accepted my answer for part (d).