A very long straight wire carries 5.0 nC on each metre of its length…

Step 1. At 0.20 m. $E = \dfrac{\lambda}{2\pi\varepsilon_0 r} = \dfrac{5.0 \times 10^{-9}\ \text{C/m}}{2\pi \times 8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2) \times 0.20\ \text{m}} = 449.6\ \text{N/C}$, about $450\ \text{N/C}$.

Step 2. At 0.50 m. The distance is $2.5$ times larger, so the field is $2.5$ times smaller: $\dfrac{449.6}{2.5} = 179.8\ \text{N/C}$, about $180\ \text{N/C}$.

Step 3. The force. $F = qE = 2.0 \times 10^{-9}\ \text{C} \times 449.6\ \text{N/C} = 8.99 \times 10^{-7}\ \text{N}$. That is $0.899\ \mu\text{N}$, directed away from the wire.