A charge of +4.0 µC and a charge of -6.0 µC are 0.12 m apart in air…

Step 1. In air. $F = \dfrac{kq_1q_2}{r^2}$. The top is $9.0 \times 10^9 \times 4.0 \times 10^{-6} \times 6.0 \times 10^{-6}\ \text{N m}^2 = 0.216\ \text{N m}^2$. The bottom is $(0.12\ \text{m})^2 = 0.0144\ \text{m}^2$. So $F = 15\ \text{N}$. The charges are unlike, so the force is an attraction.

Step 2. At twice the distance. $r^2$ becomes $4$ times larger, so $F = \dfrac{15\ \text{N}}{4} = 3.75\ \text{N}$.

Step 3. In oil. $F = \dfrac{15\ \text{N}}{2.5} = 6.0\ \text{N}$.