= Solution
A positive-parity $K=0$ <nuclear rotational band> in an <even-even nucleus> has $J=0,2,4,\ldots$ and excitation <energies>
$$
E_J=\frac{\hbar^2}{2\mathcal I}J(J+1).
$$
The four nonzero levels give $E_J/[J(J+1)]=15.17,14.85,14.48,14.01\ \mathrm{keV}$. These approximately constant coefficients, together with the even-spin positive-<parity> sequence, support a rotational interpretation. The roughly eight-percent decline means a constant-inertia rigid rotor is only an approximation: the effective <moment of inertia> increases along the sequence.
Using the lowest excited level gives
$$
\boxed{\mathcal I=\frac{3\hbar^2}{91\ \mathrm{keV}}
=2.29\times10^{-54}\ \mathrm{kg\,m^2}.}
$$
Using the highest level instead gives $2.48\times10^{-54}\ \mathrm{kg\,m^2}$, showing the scale of the departure from a constant inertia. This is a comparison of <nuclear rotational inertia relative to a rigid sphere>, not evidence for an exactly spherical rotating intrinsic state.
For a conventional size estimate, explicitly take a uniform sharp <radius> $R=r_0A^{1/3}$ with $r_0\simeq1.2\ \mathrm{fm}$ and <mass> $M\simeq A m_u$. These are standard scale approximations, not additional measured properties of this isotope supplied by the paper. They give $R\simeq6.70\ \mathrm{fm}$ and
$$
\mathcal I_{\rm rigid}=\frac25MR^2\simeq5.19\times10^{-54}\ \mathrm{kg\,m^2},\qquad
\boxed{\frac{\mathcal I}{\mathcal I_{\rm rigid}}\simeq0.44
\left(\frac{1.2\ \mathrm{fm}}{r_0}\right)^2.}
$$
Thus the fitted rotational inertia is substantially smaller than the rigid-body scale. <Nuclear pairing> provides a physical reason collective nuclear inertia need not equal the classical rigid-body value. The comparison's numerical <radius> dependence is explicit; it is not an independently determined <radius>.
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