The eigenvectors of the Hamiltonian will be the same as the eigenvectors of $ \hat{L}^2 $, and their eigenvalues will be the eigenvalues of $ \hat{L}^2 $ divided by $ 2I $. The eigenvectors of $ \hat{L}^2 $ are the $ \left | lm \right \rangle $ states (which are also eigenvectors of $ \hat{L}_z $). The eigenvalues will be $ \frac{1}{2I} \hbar^2 l \left ( l + 1 \right ) $ and the degeneracy for each value of $ l $ will be the number of possible values of $ m $ - namely, $ 2l + 1 $.
Perhaps check out this wikipedia section. If you want "explicit" states, check out this other wikipedia section and the spherical harmonics.
Proof:
Let's say $ \left | lm \right \rangle $ is an eigenvector of $ \hat{L}^2 $. We see:
\begin{align}
\hat{H} \left | lm \right \rangle & = \frac{1}{2I} \hat{L}^2 \left | lm \right \rangle = \frac{1}{2I} \hbar^2 l \left ( l + 1 \right ) \left | lm \right \rangle
\end{align}
Therefore $ \left | lm \right \rangle $ is an eigenvector of $ \hat{H} $ with eigenvalue $ \frac{1}{2I} \hbar^2 l \left ( l + 1 \right ) $. Furthermore, if $ \left | \psi \right \rangle $ is an eigenvector of $ \hat{H} $:
\begin{align}
\hat{H} \left | \psi \right \rangle = E \left | \psi \right \rangle & = \frac{1}{2I} \hat{L}^2 \left | \psi \right \rangle \\
\hat{L}^2 \left | \psi \right \rangle & = 2 I E \left | \psi \right \rangle
\end{align}
This shows that every eigenvector of the Hamiltonian is also an eigenvector of $ \hat{L}^2 $. Therefore, each eigenvector of the Hamiltonian must be one of the $ \left | lm \right \rangle $ states.
It is in general true that operators proportional to each other share eigenvectors with eigenvalues related by the same proportionality relationship. The above proof works with $ \frac{1}{2I} $ as the general proportionality constant between two general operators.
