Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
Two identical point charges $q$ are separated by a distance of 1.5 cm. The repulsive force exerted by each one of them on the other is about 0.02 N.
Find the charge magnitude $q$.
$$\color{blue}{\bf [b]}$$
Solving the given formula for $q$;
$$q=\sqrt{\dfrac{(0.015)^2(0.02)}{(9\times 10^9)}}=\pm 2.2\times 10^{-8}\;\rm C$$
The two charges could be positive or negative.
$$q=\color{red}{\bf \pm 22}\;\rm nC$$
