A small metal ball carries a charge of -4.8 nC…

Step 1. The field. $E = \dfrac{kQ}{r^2} = \dfrac{9.0 \times 10^{9} \times 4.8 \times 10^{-9}}{(0.12)^2}\ \text{N/C} = \dfrac{43.2}{0.0144}\ \text{N/C} = 3000\ \text{N/C}$. The ball is negative, so the field points towards the ball.

Step 2. The force. $F = qE = 2.0 \times 10^{-9}\ \text{C} \times 3000\ \text{N/C} = 6.0 \times 10^{-6}\ \text{N}$, that is $6.0\ \mu\text{N}$. The charge is positive, so the force is along the field, towards the ball.

Step 3. At twice the distance. The field falls to a quarter, so $E = \dfrac{3000}{4} = 750\ \text{N/C}$.