The question is about calculating the electric field at the surface of a thin spherical shell with a uniform surface charge density denoted by $$\sigma$$. To determine the electric field at any point on the surface of the shell, we can use Gauss's law, which is particularly useful for systems with high symmetry like a spherical shell.

Gauss's law in its integral form states that the electric flux through a closed surface is equal to the charge enclosed by the surface divided by the permittivity of free space ($$ \epsilon_0 $$), mathematically represented as:

$$ \Phi_E = \frac{Q_{\text{enc}}}{\epsilon_0} $$

In the case of a spherical shell of radius $$R$$, the total charge $$Q_{\text{enc}}$$ on the shell can be written in terms of the surface charge density $$\sigma$$ as:

$$ Q_{\text{enc}} = \sigma \times 4\pi R^2 $$

Now, we apply Gauss's law using a Gaussian surface that coincides with the surface of the spherical shell. The electric field $$E$$ at the surface of the shell is uniform over the Gaussian surface, and the area of the Gaussian surface (which is also the area of the spherical shell) is $$4\pi R^2$$. Thus, the electric flux $$\Phi_E$$ through the Gaussian surface is:

$$ \Phi_E = E \cdot 4\pi R^2 $$

Using Gauss's law:

$$ E \cdot 4\pi R^2 = \frac{\sigma \cdot 4\pi R^2}{\epsilon_0} $$

Simplifying this equation gives us the electric field at the surface of the spherical shell:

$$ E = \frac{\sigma}{\epsilon_0} $$

Therefore, the correct answer is Option B:

$$ \frac{\sigma}{\epsilon_0} $$