Step 1: Set up the density formula for a fcc metal.
For any fcc unit cell, density is $\rho = \dfrac{ZM}{a^3 N_A}$, where $Z = 4$ for fcc, $M$ is the molar mass and $a$ is the edge length. Since $Z$ and $N_A$ are the same constant for both metals, they will cancel when we take a ratio, so we do not need their actual values.
Step 2: Write the ratio the way the question asks for it, Cu to Ag, from the start.
\[ \frac{\rho_{Cu}}{\rho_{Ag}} = \frac{M_{Cu}}{M_{Ag}} \times \frac{a_{Ag}^3}{a_{Cu}^3} \]
Setting it up this way avoids having to flip the ratio at the end.
Step 3: Use the given edge length ratio.
We are told $a_{Cu} : a_{Ag} = 9 : 10$, so $\dfrac{a_{Ag}}{a_{Cu}} = \dfrac{10}{9}$.
\[ \left(\frac{a_{Ag}}{a_{Cu}}\right)^3 = \left(\frac{10}{9}\right)^3 = \frac{1000}{729} \approx 1.372 \]
Step 4: Use the given molar masses.
\[ \frac{M_{Cu}}{M_{Ag}} = \frac{63.5}{108} \approx 0.588 \]
Step 5: Multiply the two factors.
\[ \frac{\rho_{Cu}}{\rho_{Ag}} = 0.588 \times 1.372 \approx 0.807 \]
This rounds to about $0.8$, matching one option cleanly, so no rounding ambiguity is left.
Step 6: Final answer.
\[ \boxed{0.8} \]