= Solution
For cubes of side $\ell$, total area divided by total mass is $S=6/(\rho\ell)$, independently of how many cubes share the $100\,\mathrm g$ sample. With $S=5000\,\mathrm{m^2\,kg^{-1}}$ and $\rho=2330\,\mathrm{kg\,m^{-3}}$,
$$
\boxed{\ell=\frac6{\rho S}=5.15\times10^{-7}\,\mathrm m.}
$$
This assumes accessible external faces, negligible contact-area losses and equally sized nonporous cubes; <BET adsorption> can also count accessible internal surface in a porous powder.
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