Two identical charged particles each having a mass 10g and charge 2...

Normal reaction applied by table on object will be, $N = mg$.

Both charges are same, hence a repulsive force will act between them.

Value of force acting will be, $\dfrac{k{Q}^{2}}{{L}^{2}}$.

Now the maximum value of friction force which table can provide is,

$= \muN = \mumg$

For equilibrium, $\dfrac{k{Q}^{2}}{{L}^{2}} = \mumg$.

$L = \sqrt{\dfrac{k{Q}^{2}}{\mumg}} = \sqrt{\dfrac{k}{0 . 25mg}}Q = 2\sqrt{\dfrac{k}{mg}}Q$

$= 2\sqrt{\dfrac{9 \times {10}^{9}}{10 \times {10}^{- 3} \times 10}}\left(2 \times {10}^{- 7}\right) = 0 . 12m = 12cm$