Two different stable nuclei have their nuclear radii in the ratio 3...

The nuclear radius $R$ is related to the mass number $A$ by the formula $R = R_0 A^{1/3}$.

Let $R_1$ and $R_2$ be the radii of the heavier and lighter nuclei respectively, and $A_1$ and $A_2$ be their mass numbers.

Given $\dfrac{R_1}{R_2} = \dfrac{3}{2}$ and $A_1 = 162$.

Using the relation, we have $\dfrac{R_1}{R_2} = \left(\dfrac{A_1}{A_2}\right)^{1/3}$

$\dfrac{3}{2} = \left(\dfrac{162}{A_2}\right)^{1/3}$

Cubing both sides, we get:

$\dfrac{27}{8} = \dfrac{162}{A_2}$

$A_2 = \dfrac{162 \times 8}{27}$

$A_2 = 6 \times 8 = 48$

The mass number of the lighter nucleus is $48$.

Answer: $48$