1. Given: A thin spherical shell of radius $R$ has uniform surface charge density $\sigma$.

2. First find total charge on the shell: $Q = \sigma \cdot 4\pi R^2$

3. Electric field due to a uniformly charged spherical shell:

   • For points outside the shell, it behaves like a point charge at the center: $E = \frac{1}{4\pi\epsilon_0}\frac{Q}{r^2}$
   • For points inside the shell: $E=0$

4. At the surface $r=R$: $E_{\text{outside at surface}} = \frac{1}{4\pi\epsilon_0}\frac{Q}{R^2}$ Substitute $Q=4\pi R^2\sigma$: $E = \frac{1}{4\pi\epsilon_0}\frac{4\pi R^2\sigma}{R^2} = \frac{\sigma}{\epsilon_0}$

5. But for a surface charge distribution, the electric field is discontinuous across the surface. The standard boundary condition is: $E_{\text{out}} - E_{\text{in}} = \frac{\sigma}{\epsilon_0}$ Here, $E_{\text{in}}=0, \qquad E_{\text{out}}=\frac{\sigma}{\epsilon_0}$

6. Field exactly on the surface: For a thin charged surface, the field at the surface is taken as the average of the fields just outside and just inside: $E_{\text{surface}} = \frac{E_{\text{out}}+E_{\text{in}}}{2} = \frac{\sigma/\epsilon_0 + 0}{2} = \frac{\sigma}{2\epsilon_0}$

7. Check options:

   • A: $\sigma/(\epsilon_0 R)$ ❌ wrong dimensions
   • B: $\sigma/\epsilon_0$ ❌ this is just outside the surface, not exactly on the charged surface
   • C: $\sigma/(2\epsilon_0)$ ✅ correct
   • D: $\sigma/(4\epsilon_0)$ ❌ incorrect

Final answer: $\boxed{\frac{\sigma}{2\epsilon_0}}$ So the correct option is C.