 Solution

Interpreting the nominal size as diameter $d=10\mathrm{nm}$, fluid viscous dominance requireswith ${\rho }_f$ in $\mathrm{kg}{\mathrm{m}}^{-3}$. The liquid density is not supplied, so the conventional fluid-Reynolds crossover is not numerically fixed. As an explicitly additional illustrative assumption, ${\rho }_f=870\mathrm{kg}{\mathrm{m}}^{-3}$ gives $\mathrm{Re}=1$ near $69\mathrm{m}{\mathrm{s}}^{-1}$; ordinary much slower nanoparticle motion is well within the viscous regime.

If instead the supplied gold density is inserted into the dimensionless ratio ${\rho }_{p}Ud/\eta$, that ratio reaches one at $3.0\mathrm{m}{\mathrm{s}}^{-1}$. It is not the fluid Reynolds number. Including the spherical Stokes prefactor, the particle relaxation criterion ${\tau }_{p}U/d=1$ occurs at $54\mathrm{m}{\mathrm{s}}^{-1}$. These quantify different inertia comparisons. The wording's speed direction is reversed: viscous dominance is obtained by making speed sufficiently small. If 10 nm denotes radius, the length-based thresholds halve.