Answer
(a) $E = 0$
(b) $E = 1.22 \times 10^{8} \mathrm{N} / \mathrm{C} $
(c) $E = 3.64 \times 10^{7} \mathrm{N} / \mathrm{C} $
Work Step by Step
(a) The electric field inside any conductor is zero due to no excess charge inside the conductor. So, the electric field inside the sphere is zero
$$E = 0$$
(b) Outside the paint layer means at distance equal to the radius of the sphere $r = 12 cm /2 = 6 cm = 0.06 m$ and the elctric field at this disatnce is given by
\begin{aligned}
|E |&=\frac{1}{4 \pi \epsilon_{0}} \frac{|q|}{r^{2}} \\ &=\frac{1}{4 \pi \epsilon_{0}} \frac{\left|-49 \times 10^{-6} \mathrm{C}\right|}{(0.06 \mathrm{m})^{2}} \\ &=1.22 \times 10^{8} \mathrm{N} / \mathrm{C}
\end{aligned}
(c) The distance outside the sphere 5 cm will be added to the radius of the sphere and the new distance where the electric field is measured will be $r$ = 5 cm +6 cm = 11 cm = 0.11 m and the electric field will be
\begin{aligned}
|E| &=\frac{1}{4 \pi \epsilon_{0}} \frac{|q|}{r^{2}} \\
&=\frac{1}{4 \pi \epsilon_{0}} \frac{ |-49 \times 10^{-6} \mathrm{C}|}{(0.11 \mathrm{m})^{2}} \\
&=3.64 \times 10^{7} \mathrm{N} / \mathrm{C}
\end{aligned}
