Answer
(a) $E = 7.44 \mathrm{N} / \mathrm{C} $
(b) $E = 0$
Work Step by Step
(a) The electric field due to this charge is given by
$$ E=\frac{1}{4 \pi \epsilon_{o}} \frac{q}{r^{2}}=9 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2} \frac{\left(2.5 \times 10^{-10} \mathrm{C}\right)}{(0.55 \mathrm{m})^{2}}=7.44 \mathrm{N} / \mathrm{C} $$
(b) Inside the conductor there are no excess charges, so the electric field is zero $$E = 0 $$
