When two conducting spheres are in contact, they achieve the same electric potential. Consider two metal spheres with radii $ R $ and $ 3R $ that have the same surface charge density $\sigma$.

For a conductive sphere, the potential $ V $ of a sphere can be expressed as $ V = \frac{\sigma r}{\varepsilon_0} $, where $ r $ is the radius of the sphere and $\varepsilon_0$ is the permittivity of free space.

When the two spheres are brought into contact and then separated, they equalize their potentials:

$ V_1 = V_2 $

Thus, the equations become:

$ \sigma_1 r_1 = \sigma_2 r_2 $

Substituting for the radii:

$ \sigma_1 R = \sigma_2 (3R) $

From this, we can solve for the ratio of the surface charge densities:

$ \frac{\sigma_1}{\sigma_2} = \frac{3R}{R} = 3 $

Therefore, the ratio of the surface charge densities on the smaller sphere to the larger sphere after contact is 3.