The correct answer is C. 36 x 108 N.
The electrostatic force between two charges is given by the equation:
$$F = \frac{kq_1q_2}{r^2}$$
where $k$ is the Coulomb constant, $q_1$ and $q_2$ are the charges, and $r$ is the distance between the charges.
In this case, $k = 8.9875517923 \times 10^9 \frac{N \cdot m^2}{C^2}$, $q_1 = q_2 = 1 \text{ C}$, and $r = 0.5 \text{ m}$. Substituting these values into the equation, we get:
$$F = \frac{(8.9875517923 \times 10^9 \frac{N \cdot m^2}{C^2})(1 \text{ C})(1 \text{ C})}{(0.5 \text{ m})^2} = 36 \times 10^8 \text{ N}$$
Therefore, the electrostatic force between two charges of one coulomb each and placed at a distance of 0.5 m will be 36 x 108 N.
Option A is incorrect because it is too small. Option B is incorrect because it is too large. Option D is incorrect because it is too small.