A thin plastic rod is bent into a half circle of radius 0.10 m…

Step 1. The density. The length of the half circle is $\pi R$, so $\lambda = \dfrac{Q}{\pi R} = \dfrac{6.0\ \text{nC}}{\pi \times 0.10\ \text{m}} = 19.1\ \text{nC/m}$.

Step 2. One piece. A piece at an angle $\varphi$ from the line of symmetry carries $dq = \lambda R\,d\varphi$. At the centre it makes $dE = \dfrac{k\,dq}{R^2} = \dfrac{k\lambda}{R}\,d\varphi$, pointing away from the piece. Its part along the line of symmetry is $dE\cos\varphi$. The part across it cancels with the piece at $-\varphi$.

Step 3. Add. $E = \dfrac{k\lambda}{R}\int_{-\pi/2}^{\pi/2}\cos\varphi\,d\varphi = \dfrac{2k\lambda}{R} = \dfrac{2kQ}{\pi R^2} = \dfrac{2 \times 9.0 \times 10^{9} \times 6.0 \times 10^{-9}}{\pi \times 0.010}\ \text{N/C} = 3438\ \text{N/C}$.

Step 4. The comparison. A point charge of $6.0$ nC at $0.10$ m gives $\dfrac{kQ}{R^2} = 5400\ \text{N/C}$. The ratio is $\dfrac{3438}{5400} = 0.637$, which is $\dfrac{2}{\pi}$.