= Solution
The rotational <kinetic energy> is $K=L^2/(2I)$. For a three-dimensional <linear rigid rotor>, <angular momentum operator> replaces the squared magnitude by $\hbar^2l(l+1)$, where $l=0,1,2,\ldots$. Hence
$$
E_l=\frac{\hbar^2}{2I}l(l+1),\qquad \frac{\hbar^2}{I}=1.1121\times10^{-21}\,\mathrm J=6.941\times10^{-3}\,\mathrm{eV}.
$$
The three lowest distinct <energy levels> are therefore
$$
\boxed{E_0=0,\qquad E_1=6.94\times10^{-3}\,\mathrm{eV},\qquad E_2=2.08\times10^{-2}\,\mathrm{eV}.}
$$
Their magnetic <quantum degeneracy> values are $1,3,5$. A single moment of inertia specifies this <rigid rotor> model; an arbitrary asymmetric body would require its other principal moments.
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