Given:

• Atomic substance A with molar mass $M = 12 \, \text{g mol}^{-1}$


• Cubic crystal structure with edge length $a = 300 \, \text{pm} = 300 \times 10^{-12} \, \text{m}$


• Density $\rho = 3.0 \, \text{g mL}^{-1}$


• Avogadro's number $N_A = 6.02 \times 10^{23} \, \text{mol}^{-1}$

First, calculate the volume of the unit cell $V = a^3 = (300 \times 10^{-12} \, \text{m})^3 = 27 \times 10^{-30} \, \text{m}^3$.

Next, calculate the mass of the unit cell using the given density. Density is mass over volume, so mass

$m = \rho \cdot V = 3.0 \, \text{g mL}^{-1} \cdot 27 \times 10^{-30} \, \text{m}^3 \cdot \frac{1 \, \text{mL}}{10^{-6} \, \text{m}^3} = 81 \times 10^{-24} \, \text{g}$.

Then, calculate the number of moles in one unit cell. The molar mass is mass over number of moles, so

$n = \frac{m}{M} = \frac{81 \times 10^{-24} \, \text{g}}{12 \, \text{g mol}^{-1}} = 6.75 \times 10^{-26} \, \text{mol}$.

Finally, calculate the number of atoms in one unit cell. The Avogadro constant is the number of atoms per mole,

so number of atoms $= n \cdot N_A = 6.75 \times 10^{-26} \, \text{mol} \cdot 6.02 \times 10^{23} \, \text{mol}^{-1} = 4.06$.

Rounding to the nearest integer, we get 4 atoms. So, there are 4 atoms present in one unit cell of substance A.