A small metal ball carrying +5.0 nC sits at the centre of a hollow plastic sphere of radius 0.20 m…

Step 1. The first sphere. $\Phi = \dfrac{q}{\varepsilon_0} = \dfrac{5.0 \times 10^{-9}\ \text{C}}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 565\ \text{N m}^2/\text{C}$.

Step 2. The second sphere. The same charge is inside, so the flux is the same, $565\ \text{N m}^2/\text{C}$.

Step 3. The field. The field is radial and has the same size on the sphere. So $E = \dfrac{\Phi}{4\pi r^2} = \dfrac{565\ \text{N m}^2/\text{C}}{4\pi \times (0.20\ \text{m})^2} = 1124\ \text{N/C}$.