A thin plastic hoop of radius 0.15 m carries charge spread evenly along it, with 8.0 nC on each…

Step 1. The charge. $Q = \lambda \times 2\pi R = 8.0\ \text{nC/m} \times 2\pi \times 0.15\ \text{m} = 7.54\ \text{nC}$.

Step 2. The ring field. $R^2 + z^2 = 0.0225 + 0.040 = 0.0625\ \text{m}^2$, and $(0.0625)^{3/2} = 0.015625\ \text{m}^3$. So $E = \dfrac{9.0 \times 10^{9} \times 7.54 \times 10^{-9} \times 0.20}{0.015625}\ \text{N/C} = 869\ \text{N/C}$.

Step 3. A point charge. $E = \dfrac{kQ}{z^2} = \dfrac{9.0 \times 10^{9} \times 7.54 \times 10^{-9}}{0.040}\ \text{N/C} = 1696\ \text{N/C}$.