The distance between two point charges $q_1=5 mC$ and $q_2=-10 mC$ in vacuum so that electrostatic force of attraction between them is 50 kN is:

3 m

The correct answer is Option (2) → 3 m

$F = k \frac{|q_1 q_2|}{r^2}$

$q_1 = 5 \times 10^{-3}, \quad q_2 = 10 \times 10^{-3}$

$F = 50 \times 10^{3}, \quad k = 9 \times 10^9$

$50 \times 10^{3} = 9 \times 10^9 \cdot \frac{5 \times 10^{-3} \cdot 10 \times 10^{-3}}{r^2}$

$50 \times 10^{3} = 9 \times 10^9 \cdot \frac{50 \times 10^{-6}}{r^2}$

$50 \times 10^{3} = \frac{450 \times 10^{3}}{r^2}$

$r^2 = \frac{450 \times 10^{3}}{50 \times 10^{3}} = 9$

$r = 3 \ \text{m}$

The distance is $3 \ \text{m}$.