A straight rod 0.80 m long carries +6.4 nC spread evenly…

Step 1. The rod. $\sqrt{r^2 + (L/2)^2} = \sqrt{0.010 + 0.160}\ \text{m} = 0.4123\ \text{m}$. So $E = \dfrac{9.0 \times 10^{9} \times 6.4 \times 10^{-9}}{0.10 \times 0.4123}\ \text{N/C} = 1397\ \text{N/C}$.

Step 2. The long wire. $\lambda = \dfrac{6.4\ \text{nC}}{0.80\ \text{m}} = 8.0\ \text{nC/m}$. Then $E = \dfrac{2k\lambda}{r} = \dfrac{2 \times 9.0 \times 10^{9} \times 8.0 \times 10^{-9}}{0.10}\ \text{N/C} = 1440\ \text{N/C}$.

Step 3. The difference. $\dfrac{1440 - 1397}{1397} = 0.031$, so $3.1\ \%$.