Size of the Nucleus

Nuclear radius $R = R_0A^{1/3}$ $\frac{R_{125}}{R_{64}} = \frac{R_0(125)^{1/3}}{R_0(64)^{1/3}} = \frac{5}{4}$ So the ratio is 5 : 4.

A nucleus of mass number $A$ has radius $R = R_0 A^{1/3}$, where $R_0 = 1.2 \times 10^{-15}$ m.
For $^{27}Al$, let the nuclear radius be $R_1$: $R_1 = R_0 (27)^{1/3} = 3R_0$
For $^{125}Te$, let the nuclear radius be $R_2$: $R_2 = R_0 (125)^{1/3} = 5R_0$
$\dfrac{R_2}{R_1} = \dfrac{5R_0}{3R_0} = \dfrac{5}{3}$
$R_2 = \dfrac{5}{3} \times R_1 = \dfrac{5}{3} \times 3.6$
$R_2 = 6$ fm

Nuclear radius: $R = R_0 A^{1/3}$, so $A \propto R^3$.
$\dfrac{A_{Ge}}{A_{Be}} = \left(\dfrac{R_{Ge}}{R_{Be}}\right)^3$
$\Rightarrow A_{Ge} = \left(\dfrac{R_{Ge}}{R_{Be}}\right)^3 (A_{Be}) = (2)^3 (9)$
$A_{Ge} = 8 \times 9 = 72$
So the number of nucleons in Ge is 72.

Size of an atom $\approx 10^{-10}$ m; size of a nucleus $\approx 10^{-15}$ m.
Volume is proportional to the cube of the radius.
$\dfrac{\text{Volume of atom}}{\text{Volume of nucleus}} \sim \left(\dfrac{10^{-10}}{10^{-15}}\right)^3 = (10^5)^3 = 10^{15}$
So the atom occupies about $10^{15}$ times the volume of its nucleus; the atom is mostly empty space.