= Solution
Assuming additive liquid volumes, the scattering length density is linear in D$_2$O volume fraction $\phi_D$:
$$
2.080=-0.560(1-\phi_D)+6.335\phi_D,
\qquad \phi_D=0.3828,
$$
where all scattering length densities are in $10^{-6}\,\mathrm{\mathring A^{-2}}$. The $100\,\mathrm g$ of heavy water occupies about $90.99\,\mathrm{cm^3}$. The required ordinary-water volume is
$$
V_H=V_D\frac{1-\phi_D}{\phi_D}\simeq146.7\,\mathrm{cm^3}.
$$
Thus add approximately \b[$147\,\mathrm g$ of H$_2$O]. Volume fraction, rather than mass fraction, enters this contrast-matching equation. Isotope exchange to form HDO does not change the total nuclear scattering-length balance in the additive-volume approximation.
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