Consider the mass of iron nucleus as 55.85 u and A = 56. Then the nuclear density is:

Correct option is D

​$$\text{Given:} \\\text{Mass of iron nucleus } m = 55.85 \, \text{u} \\\text{Atomic mass unit } u = 1.66 \times 10^{-27} \, \text{kg} \\\text{Atomic number } A = 56 \\\text{Radius of nucleus } R_0 = 1.2 \times 10^{-15} \, \text{m}$$

$$\text{The volume of the nucleus is given by:} \\V = \dfrac{4}{3} \pi R_0^3 A$$

$$\text{Substitute the known values:} \\V = \dfrac{4}{3} \pi (1.2 \times 10^{-15})^3 \times 56$$

$$\text{The nuclear density is the mass per unit volume:} \\\rho_N = \dfrac{M}{V}$$

​$$\text{Where:} \\M = 55.85 \times 1.66 \times 10^{-27} \, \text{kg }$$

$$\text{Substitute the values into the equation:} \\\rho_N = \dfrac{55.85 \times 1.66 \times 10^{-27}}{\dfrac{4}{3} \pi (1.2 \times 10^{-15})^3 \times 56}$$

$$\rho_N = 2.29 \times 10^{17} \, \text{kg/m}^3$$​​