Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The Rutherford atom is neutral, the nucleus has a net charge of $Q_n=+Ze$ and the electron cloud has a net charge of $Q_e=-Ze$ where $Z$ is the atomic number.
As the author told us, we can assume that the charge is uniformly distributed inside the electron's cloud.
This means that the volume charge density is constant everywhere
$$\rho_e=\dfrac{Q_{\rm net}}{V}=\dfrac{-Ze}{\frac{4}{3}\pi R^3}\tag 1$$
For the Gaussian sphere, we know that
$$\oint \vec Ed\vec A=\dfrac{Q_{in}}{\epsilon_0}$$
And since the charge is uniformly distributed, the electric field is uniform. So,
$$EA=\dfrac{Q_{in}}{\epsilon_0}$$
Thus,
$$E =\dfrac{Q_{in}}{A\epsilon_0}$$
where $A=4\pi r^2$ where $r$ is the radius of the Gaussian sphere.
$$E =\dfrac{Q_{in}}{4\pi r^2\epsilon_0}\tag 2$$
Now we need to find $Q_{in}$,
$$Q_{in}=Q'_e+Q_n$$
where the enclosed charge is the whole positive charge of the nucleus $Q_n$ plus some of the electrons $Q'_e$.
$$Q_{in}=Q'_e+Ze\tag 3$$
Now we need to find $Q'_e$ by using (1),
$$\rho_e =\dfrac{-Ze}{\frac{4}{3}\pi R^3}=\dfrac{Q'_e}{\frac{4}{3}\pi r^3}$$
where $r\lt R$,
Hence,
$$Q'_e=\dfrac{-Ze r^3} { R^3}$$
Plug into (3),
$$Q_{in}=\dfrac{-Ze r^3} { R^3}+Ze=Ze\left[ 1-\dfrac{ r^3} { R^3}\right]$$
Plug into (2),
$$E =\dfrac{Ze}{4\pi r^2\epsilon_0}\left[ 1-\dfrac{ r^3} {R^3}\right]$$
Therefore,
$$\boxed{E =\dfrac{Ze}{4\pi \epsilon_0}\left[ \dfrac{1}{r^2}-\dfrac{ r} {R^3}\right]}$$
And since the positive charge is greater than the negative charge, the direction of the electric field is outward.
$$\color{blue}{\bf [b]}$$
On the surface of atom $r=R$, so
$$E =\dfrac{Ze}{4\pi \epsilon_0}\left[ \dfrac{1}{R^2}-\dfrac{ R} {R^3}\right]=\dfrac{Ze}{4\pi \epsilon_0}\left[ \dfrac{1}{R^2}-\dfrac{ 1} {R^2}\right]$$
$$E =\color{red}{\bf 0}\;\rm N/C$$
which is a reasonable result since the atom as one unit is neutral and has a net charge of zero. So the enclosed charge then is zero, and hence the electric field is zero.
$$\color{blue}{\bf [c]}$$
Plugging the known into the boxed formula from part (a) above,
$$ E =\dfrac{Ze}{4\pi \epsilon_0}\left[ \dfrac{1}{(\frac{1}{2}R)^2}-\dfrac{ \frac{1}{2}R} {R^3}\right] $$
$$ E =\dfrac{Ze}{4\pi \epsilon_0}\left[ \dfrac{4}{ R^2}-\dfrac{1} {2R^2}\right] $$
$$ E =\dfrac{Ze}{4\pi \epsilon_0} \dfrac{7}{2 R^2} $$
$$ E = (9\times 10^9)(92)(1.6\times 10^{-19}) \dfrac{7}{2 (0.1\times 10^{-9})^2} $$
$$E=\color{red}{\bf 4.64\times10^{13}}\;\rm N/C$$
