A point charge of +4.0 nC is placed on a line through the centre of a sphere of radius 0.30 m…

Step 1. The comparison number. $\dfrac{q}{\varepsilon_0} = \dfrac{4.0 \times 10^{-9}\ \text{C}}{8.85 \times 10^{-12}\ \text{C}^2/(\text{N m}^2)} = 452\ \text{N m}^2/\text{C}$.

Step 2. $d = 0.10$ m. This is less than the radius $0.30$ m, so the charge is inside. The net flux is $\dfrac{q}{\varepsilon_0} = 452\ \text{N m}^2/\text{C}$, wherever the charge sits inside.

Step 3. $d = 0.60$ m. The charge is outside. Every line that enters the sphere leaves again, so the net flux is $0\ \text{N m}^2/\text{C}$.