Answer
$2.4\times10^{-10}N$.
Work Step by Step
There are four forces to calculate; we ignore the two “internal” forces. Let the repulsive and attractive forces have negative and positive signs, respectively.
$$F_{net}=F_{CH}+F_{CN}+F_{OH}+F_{ON}$$
$$ =\frac{k(0.40e)(0.20e)}{(1\times10^{-9}m)^2}(-\frac{1}{(0.30)^2}+\frac{1}{(0.40)^2}+\frac{1}{(0.18)^2}-\frac{1}{(0.28)^2}) $$
$$ =2.4\times10^{-10}N$$
