Answer
(a) $\Phi_{E} = 0 $
(b) $\Phi_{E} = -678 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$
(c) $\Phi_{E} = -226 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}$
Work Step by Step
(a) Inside the conductor, there are no excess charges so the electric flux is zero
$$ \Phi_{E}=0 $$
(b) Inside radius 1.5 cm, the sphere contains $q_2$ only, so the electric flux will be
$$ \Phi_{E}=\frac{q_{2}}{\epsilon_{0}}=\frac{-6.00 \times 10^{-9} \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}}=-678 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} $$
(c) Inside radius 2.5 cm, the sphere contains charges $q_1$ and $q_2$ and the electric flux will be
$$ \Phi_{E}=\frac{q_{1}+q_{2}}{\epsilon_{0}}=\frac{(4.00-6.00) \times 10^{-9} \mathrm{C}}{8.85 \times 10^{-12} \mathrm{C}^{2} / \mathrm{N} \cdot \mathrm{m}^{2}}=-226 \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C} $$