The radius R of a nucleus of mass number A can be estimated by the formula $$R = (1.3 \times 10^{-15})A^{1/3}$$ m. It follows that the mass density of n nucleus is of the order of: $$(M_{prot} \cong M_{neut} \simeq 1.67 \times 10^{-27}$$ kg)

We start with the general definition of mass density. Density $$\rho$$ is the ratio of mass $$M$$ to volume $$V$$:

$$\rho \;=\;\dfrac{M}{V}$$

For a nucleus having mass number $$A$$ (that is, a total of $$A$$ nucleons), the mass can be written by taking each nucleon (proton or neutron) to have approximately the same mass $$m_{n}$$:

$$M \;=\; A\,m_{n}$$

The data supplied give the numerical value of a nucleon’s mass:

$$m_{n} \;=\;1.67\times10^{-27}\,\text{kg}$$

Next, we use the empirical formula that links the nuclear radius $$R$$ with the mass number $$A$$:

$$R \;=\;(1.3\times10^{-15})\,A^{1/3}\;\text{m}$$

To find the volume of the nucleus we substitute this expression for $$R$$ into the standard volume formula for a sphere, $$V=\dfrac{4}{3}\pi R^{3}$$ :

$$V \;=\;\dfrac{4}{3}\,\pi\left[(1.3\times10^{-15})\,A^{1/3}\right]^3$$

Now we expand the cube term by term. First, the numerical factor:

$$1.3^{3}=1.3\times1.3\times1.3=2.197$$

Second, the power of ten:

$$(10^{-15})^{3}=10^{-45}$$

Third, the power of $$A$$:

$$(A^{1/3})^{3}=A$$

Putting these pieces together yields:

$$V \;=\;\dfrac{4}{3}\,\pi\;\bigl(2.197\bigr)\times10^{-45}\;A$$

We can combine the constants $$\dfrac{4}{3}\pi$$ and $$2.197$$ into one numerical factor:

$$\dfrac{4}{3}\pi \;=\;4.18879$$

Multiplying this by $$2.197$$ gives

$$4.18879\times2.197\;\approx\;9.20$$

Therefore the volume becomes

$$V \;\approx\;9.20\times10^{-45}\;A\;\text{m}^{3}$$

Having explicit expressions for both $$M$$ and $$V$$, we substitute them into the density formula:

$$\rho \;=\;\dfrac{A\,m_{n}}{9.20\times10^{-45}\;A}$$

Notice that the factor $$A$$ cancels out, leaving the density independent of the mass number:

$$\rho \;=\;\dfrac{m_{n}}{9.20\times10^{-45}}$$

We now insert the value of $$m_{n}$$ :

$$\rho \;=\;\dfrac{1.67\times10^{-27}}{9.20\times10^{-45}}$$

To divide these numbers, we handle the mantissas and the powers of ten separately.

The mantissas: $$\dfrac{1.67}{9.20}=0.1815$$ (approximately).

The powers of ten: $$10^{-27}\div10^{-45}=10^{18}$$ because subtracting the exponents gives $$-27-(-45)=18$$.

Putting these pieces together we obtain

$$\rho\;\approx\;0.1815\times10^{18}\;\text{kg m}^{-3}$$

For order-of-magnitude purposes, $$0.1815\times10^{18}$$ can be written as $$1.8\times10^{17}$$ :

$$\rho\;\approx\;1.8\times10^{17}\;\text{kg m}^{-3}$$

This value is of the order of $$10^{17}\;\text{kg m}^{-3}$$.

Hence, the correct answer is Option D.