= Solution
The original apparatus has two small masses at distance $L/2$ from the suspension axis. Neglecting the rod's <moment of inertia>, their total <moment of inertia> is $I=2m(L/2)^2=mL^2/2$, consistent with the supplied <torsion pendulum> period. Treat $m$ and $L$ as exact because no uncertainties are specified. Rearranging the period relation gives
$$
\kappa=\frac{2\pi^2mL^2}{T^2}
=\frac{2\pi^2(0.038\,\mathrm{kg})(0.100\,\mathrm m)^2}{(498.2\,\mathrm s)^2}
=3.02208\times10^{-8}\,\mathrm{N\,m\,rad^{-1}}.
$$
The <torsion constant> is a restoring <torque> per unit <angle>. In dimensional calculations the radian is dimensionless.
Using first-order <propagation of uncertainty>, with the quoted period uncertainty interpreted as a <standard uncertainty>,
$$
\sigma_\kappa=\left|\frac{d\kappa}{dT}\right|\sigma_T
=\frac{2\kappa}{T}\sigma_T,
\qquad \frac{\sigma_\kappa}{\kappa}=\frac{2(6.0)}{498.2}=0.02409.
$$
Consequently
$$
\boxed{\kappa=(3.022\pm0.073)\times10^{-8}\,\mathrm{N\,m\,rad^{-1}}.}
$$
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