B. 3.0 km

Using the formula for the electric field due to a point charge:
\(E = \frac{1}{4 \pi \epsilon_0} \frac{q}{r^2}\)

Given: \(E = 4.8 \times 10^{-4} \, \text{N/C}, \quad q = 1.2 \times 10^{-7} \, \text{C}, \quad \frac{1}{4 \pi \epsilon_0} = 9.0 \times 10^9 \, \text{Nm}^2/\text{C}^2\)

Rearranging:
\(r^2 = \frac{(9.0 \times 10^9) \times (1.2 \times 10^{-7})}{4.8 \times 10^{-4}} = 9 \times 10^6\)

Thus, \(r = \sqrt{9 \times 10^6} = 3000 \, \text{m} = 3.0 \, \text{km}\)