The ratio of mass densities of nuclei of $$^{40}$$Ca and $$^{16}$$O is close to:

We have to compare the mass densities of two atomic nuclei, $$^{40}\text{Ca}$$ and $$^{16}\text{O}$$. Mass density (symbol $$\rho$$) is defined by the general formula $$\rho=\dfrac{\text{mass}}{\text{volume}}$$.

Inside a nucleus, almost the entire mass is contributed by the nucleons (protons + neutrons). For a nucleus whose mass number is $$A$$, the total mass is very well approximated by $$M \approx A\,m_N$$, where $$m_N$$ is the average mass of one nucleon. We shall keep $$m_N$$ common for both nuclei.

Next, we need the volume of a nucleus. Experimentally, nuclear radii follow the empirical relation

$$R = R_0\,A^{1/3},$$

where $$R_0$$ is a constant (about $$1.2\;\text{fm}$$). A nucleus is essentially spherical, so its volume is

$$V = \dfrac{4}{3}\pi R^3.$$

Substituting $$R = R_0 A^{1/3}$$, we get

$$V = \dfrac{4}{3}\pi (R_0 A^{1/3})^3 = \dfrac{4}{3}\pi R_0^{\,3}\,A.$$

Observe that the volume is directly proportional to $$A$$. Now, putting the mass and volume expressions into the density formula, we have

$$\rho = \dfrac{M}{V} = \dfrac{A\,m_N}{\dfrac{4}{3}\pi R_0^{\,3} A} = \dfrac{m_N}{\dfrac{4}{3}\pi R_0^{\,3}}.$$

Crucially, the factor $$A$$ cancels out completely. Therefore, the density $$\rho$$ is independent of the mass number; it is the same for any nucleus as long as we employ the same constant $$R_0$$ and the same nucleon mass $$m_N$$.

Now we form the required ratio:

$$\dfrac{\rho(^{40}\text{Ca})}{\rho(^{16}\text{O})} = \dfrac{m_N / \left(\dfrac{4}{3}\pi R_0^{\,3}\right)} {m_N / \left(\dfrac{4}{3}\pi R_0^{\,3}\right)} = 1.$$

So the two nuclei possess practically the same mass density. Hence, the correct answer is Option A.