Answer
See the detailed answer below.
Work Step by Step
We know that the electric field at the surface of a charged conductor is given by
$$E=\dfrac{\eta}{\epsilon_0}\tag{perpendicular to surface}$$
$\bullet$ At point 1, at the surface of this charged conductor, the electric field is
$$E_1=\dfrac{\eta}{\epsilon_0} =\dfrac{\eta q_e}{\epsilon_0}$$
where $q_e$ is the charge of one electron.
Plugging the known,
$$E_1=\dfrac{(5\times 10^{10})(-1.6\times 10^{-19})}{(8.85\times 10^{-12})} $$
$$E_1=\color{red}{\bf -904}\;\rm N/C$$
$\bullet\bullet$ At point 2, at the point which is inside the conductor, the electric field is zero.
$$E_2=\color{red}{\bf 0}\;\rm N/C$$
$\bullet\bullet\bullet$ At point 3, at the point which is inside the hollow of the conductor, the electric field is zero since the excess charge inside the conductor always resides on the outer surface.
Hence, the inner surface has no charge. Thus,
$$E_3=\color{red}{\bf 0}\;\rm N/C$$
