A heavy nucleus of mass number 64 hypothetically breaks into 8...

The radius of a nucleus is given by $R = R_0 A^{1/3}$.

The surface area of a nucleus is $S = 4\pi R^2$.

Substituting the expression for $R$, we get $S \propto A^{2/3}$.

For the original nucleus ($A = 64$), the surface area is:

$S_0 \propto (64)^{2/3} = 16$

For one lighter nucleus ($A' = 8$), the surface area is:

$S_1 \propto (8)^{2/3} = 4$

The total surface area of all $8$ lighter nuclei is:

$S_{\text{total}} = 8 \times S_1 \propto 8 \times 4 = 32$

The required ratio is:

$\dfrac{S_{\text{total}}}{S_0} = \dfrac{32}{16} = \dfrac{2}{1}$

Thus, the ratio is $2:1$.

Answer: $2:1$