An unknown nucleus has a nuclear density of 2.29 × 10^ 17 kg/m ^3...

The nuclear density $\rho$ is given by the ratio of the mass of the nucleus to its volume:

$\rho = \dfrac{m}{\dfrac{4}{3}\pi R^3}$

The radius of the nucleus is $R = R_0 A^{1/3}$, which gives $R^3 = R_0^3 A$.

Substituting this into the density formula:

$\rho = \dfrac{3m}{4\pi R_0^3 A}$

Rearranging for the mass number $A$:

$A = \dfrac{3m}{4\pi R_0^3 \rho}$

Substituting the given values:

$A = \dfrac{3 \times 19.926 \times 10^{-27}}{12.56 \times (1.2 \times 10^{-15})^3 \times 2.29 \times 10^{17}}$

$A = \dfrac{59.778 \times 10^{-27}}{12.56 \times 1.728 \times 10^{-45} \times 2.29 \times 10^{17}}$

$A = \dfrac{59.778 \times 10^{-27}}{49.7 \times 10^{-28}}$

$A = \dfrac{597.78}{49.7} \approx 12$

Answer: $12$