An unstable heavy nucleus at rest breaks into two nuclei which move away with velocities in the ratio of 8 : 27. The ratio of the radii of the nuclei (assumed to be spherical) is:

We have an original heavy nucleus that is initially at rest, so its total linear momentum is zero.

After disintegration it breaks into two daughter nuclei. Let their masses be $$m_1$$ and $$m_2$$, and let their speeds be $$v_1$$ and $$v_2$$ respectively. The problem states that the magnitudes of the velocities are in the ratio $$v_1 : v_2 = 8 : 27$$.

Because no external force acts on the system, linear momentum is conserved. Stating the conservation of linear momentum for the magnitudes, we write

$$m_1 v_1 = m_2 v_2.$$

Re-arranging,

$$\frac{m_1}{m_2} = \frac{v_2}{v_1}.$$

Substituting the given ratio $$v_1 : v_2 = 8 : 27$$, we get

$$\frac{m_1}{m_2} = \frac{27}{8}.$$

Now, each daughter nucleus is assumed to be spherical and made of the same nuclear matter, so the density $$\rho$$ is the same for both. The mass of a sphere is given by

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

where $$R$$ is its radius. Therefore mass is directly proportional to the cube of the radius:

$$\frac{m_1}{m_2} = \frac{R_1^{3}}{R_2^{3}}.$$

Substituting the previously obtained mass ratio,

$$\frac{R_1^{3}}{R_2^{3}} = \frac{27}{8}.$$

Taking the cube root of both sides,

$$\frac{R_1}{R_2} = \sqrt[3]{\frac{27}{8}} = \frac{3}{2}.$$

So the radii are in the ratio $$3 : 2$$.

Hence, the correct answer is Option C.