Answer
(a) $E = 2.4 \times 10^{21} \mathrm{N} / \mathrm{C}$
(b) $E = 1.3 \times 10^{13} \mathrm{N} / \mathrm{C}$
(c) $E = 0$
Work Step by Step
(a) The electric field inside the nucleus due to 92 protons is given by
\begin{align}
E&=\frac{1}{4 \pi \epsilon_{o} } \frac{|q|}{r^{2}}=\\
& = \frac{1}{4 \pi \epsilon_{o} } \frac{92 e}{r^{2}} \\
& \left(8.99 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{92\left(1.60 \times 10^{-19} \mathrm{C}\right)}{\left(7.4 \times 10^{-15} \mathrm{m}\right)^{2}}\\
& =\boxed{2.4 \times 10^{21} \mathrm{N} / \mathrm{C}}
\end{align}
(b) When the $r = 1 \times 10^{-10} \mathrm{m}$, the electric field is calcauted by
\begin{align}
E&=\frac{1}{4 \pi \epsilon_{o} } \frac{|q|}{r^{2}}=\\
& = \frac{1}{4 \pi \epsilon_{o} } \frac{92 e}{r^{2}} \\
& \left(8.99 \times 10^{9} \mathrm{N} \cdot \mathrm{m}^{2} / \mathrm{C}^{2}\right) \frac{92\left(1.60 \times 10^{-19} \mathrm{C}\right)}{\left(1 \times 10^{-10} \mathrm{m}\right)^{2}}\\
& =\boxed{1.3 \times 10^{13} \mathrm{N} / \mathrm{C}}
\end{align}
(c) The electrons cancel the positive charges of the protons. Hence, the electric field, in this case, equals zero.
