= Solution
Convert both radii to metres before evaluating the <moment of inertia>:
$$
I=\tfrac12m(a^2+b^2)=1.1770375\times10^{-3}\,\mathrm{kg\,m^2}.
$$
Its linear differential is
$$
dI=\tfrac12(a^2+b^2)\,dm+ma\,da+mb\,db.
$$
For <independent> <standard uncertainties>, add the three contributions in quadrature:
$$
\sigma_I=\sqrt{[\tfrac12(a^2+b^2)\sigma_m]^2+(ma\sigma_a)^2+(mb\sigma_b)^2}
=2.81\times10^{-5}\,\mathrm{kg\,m^2},
\qquad
\boxed{I=(1.177\pm0.028)\times10^{-3}\,\mathrm{kg\,m^2}.}
$$
If the quoted errors are instead bounds, their first-order absolute contributions add, giving \b[$I\simeq(1.177\pm0.042)\times10^{-3}\,\mathrm{kg\,m^2}$]. Their statistical interpretation is not specified; the two conventions should remain distinct.
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