= Solution
Use the water data to determine the common oxygen contribution rather than assume identical molar volumes. Ordinary water has molecular number density
$$
n_H=\frac{10^6}{18}N_A=3.346\times10^{28}\,\mathrm{m^{-3}},
$$
so $b_O=\rho_b^{H_2O}/n_H-2b_H\simeq5.81\times10^{-15}\,\mathrm m$. Therefore $b_{D_2O}=2b_D+b_O\simeq1.915\times10^{-14}\,\mathrm m$ and
$$
n_D=\frac{6.335\times10^{14}}{b_{D_2O}},\qquad
\boxed{\rho_{D_2O}=\frac{20n_D}{N_A}\simeq1.10\,\mathrm{g\,cm^{-3}}}.
$$
The assumptions are isotopically pure waters, additive nuclear scattering lengths, the same oxygen isotope contribution, and molar masses approximately $18$ and $20\,\mathrm{g\,mol^{-1}}$. Equal molar volumes would give the rougher estimate $20/18=1.11\,\mathrm{g\,cm^{-3}}$.
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