Answer
The distance to be increased to 3 times its initial distance.
Work Step by Step
Let $F=k\frac{q_{1}q_{2}}{d^{2}}$,
When one charge is increased by 27 times, and other reduce to one-third,
then new $F=k\frac{27q_{1}\frac{1}{3}q_{2}}{(d')^{2}}=9k\frac{q_{1}q_{2}}{(d^{'})^{2}}=k\frac{q_{1}q_{2}}{d^{2}}$,
So, $d^{'}=3d$
The distance to be increased to 3 times its initial distance.