A

Step 1: Understand that the electric flux through a closed surface is given by Gauss's Law, which states that the total electric flux, \(\Phi_E\), through a closed surface is equal to the net charge enclosed by that surface divided by the permittivity of free space, \(\epsilon_0\):

\(\Phi_E = \frac{Q_{enc}}{\epsilon_0}\)

Step 2: In this problem, the spherical surface of radius \(R\) encloses only the charge located at the center, which is \(Q = 8.85 \times 10^{-8} C\). The other charge is located at a distance of \(3R\) away from the center and does not contribute to the net charge enclosed.

Step 3: Plugging the known values into Gauss's Law:

\(\Phi_E = \frac{8.85 \times 10^{-8}}{8.85 \times 10^{-12}} = 10^{4} \, C/m^{2}\)

Step 4: Therefore, the electric flux through the spherical surface of radius \(R\) is:

\(\Phi_E = 10^{4}\, \text{N m}^2/C\)

Therefore, the answer is A.