= Solution
Write $B$ for the <filament bending modulus> and $C$ for the <filament torsional stiffness>, so that the <energy> per unit length is $(B\kappa^2+C\Omega^2)/2$. For one circular filament,
$$
B_1=EI_1=\frac{\pi Er^4}{4},\qquad C_1=G_sJ_1=\frac{\pi G_sr^4}{2},\qquad G_s=\frac{E}{2(1+\nu)},
$$
where $I_1$ is the <second moment of area>, $J_1$ the polar moment, $G_s$ the <shear modulus>, and $\nu$ <Poisson's ratio>. Numerical twisting constants cannot be fixed by <Young's modulus> alone without specifying $\nu$.
For a filled sliding bundle, $N\sim(R/r)^2$. Filaments share the centerline curvature but slide instead of accumulating the additional axial strains of a bonded cross-section. Their bending energies add: \b[$B\sim NB_1\sim Er^2R^2$]. If end constraints impose the same material twist on every filament, their torsional energies also add: \b[$C\sim G_sr^2R^2$].
For a bonded, one-filament-thick circular tube, the wall thickness $h$ is of order $r$. Its moments are $I\sim\pi R^3h$ and $J_p\sim2\pi R^3h$, giving \b[$B\sim ErR^3$ and $C\sim G_srR^3$]. For a bonded filled rod, $I\sim\pi R^4/4$ and $J_p\sim\pi R^4/2$, giving \b[$B\sim ER^4$ and $C\sim G_sR^4$]. Packing fractions and the convention for tube radius affect constants, not these powers.
There is an important qualification to <bending and torsion of filament bundles>: free axial sliding does not specify whether each filament can also spin freely. If both spin and slip are unconstrained, the common-material-twist calculation is not forced by the geometry. A helix of small twist rate has centerline curvature of order $\rho\Omega^2$, so its bending <energy> starts at $\Omega^4$, not $\Omega^2$. A nonzero linear collective torsional stiffness then requires additional end, contact, or spin constraints.
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