Answer
See the detailed answer below.
Work Step by Step
It seems that the hollow sphere is polarized since the charge density of the inner and outer surfaces are equal in charge density but opposite in signs.
This means that there must be a positive in the cavity of the hollow sphere that is equal in magnitude to the charge in the inner surface of our hollow sphere.
Now we need to find the charge magnitude in the cavity of the hollow sphere.
$$\eta=\dfrac{Q_{inner}}{A}$$
Hence,
$$Q_{\rm inner}=\eta A=4\pi r^2 \eta$$
where $r=a=6$ cm,
$$Q_{\rm inner}=4\pi a^2 \eta$$
$$Q_{\rm inner}=4\pi (0.06)^2 (-100\times 10^{-9})$$
$$Q_{\rm inner}=\bf -4.524\;\rm nC$$
Thus, the charge inside the cavity is then
$$Q_{\rm cavity}= \bf +4.524\;\rm nC$$
and hence, the charge of the outer surface of the hollow sphere is then
$$Q_{\rm outer}=4\pi b^2 \eta$$
$$Q_{\rm outer}=4\pi (0.10)^2 (100\times 10^{-9})$$
$$Q_{\rm outer}=\bf +12.57 \;\rm nC$$
In the figure below, and from the given data, we can figure out that
$a=6$ cm, $b=10$cm, $r_1=4$ cm, $r_2=8$ cm, and $r_3=12$ cm.
To find the electric field at these 3 radii, we need to use Gauss's law.
$$\oint \vec Ed\vec A=\dfrac{Q_{in}}{\epsilon_0}$$
and since $E$ is always perpendicular to the surface of our sphere and is uniform and constant,
$$EA=\dfrac{Q_{in}}{\epsilon_0}$$
where $A=4\pi r^2$
Hence,
$$\boxed{E=\dfrac{Q_{in}}{4\pi r^2\epsilon_0 }}$$
We will use this formula to find the three electric fields at the three Gaussian surfaces.
At $r_1$,
$$E_1=\dfrac{Q_{1,in}}{4\pi r_1^2\epsilon_0 }=\dfrac{Q_{\rm cavity}}{4\pi r_1^2\epsilon_0 }$$
Plug the known;
$$E_1=\dfrac{+4.524\times 10^{-9}}{4\pi (0.04)^2(8.85\times 10^{-12})}$$
$$E_1=\color{red}{\bf 2.54\times 10^4}\;\rm N/C\tag {Outward}$$
The direction of this electric field is outward since the net inner charge is positive.
At $r_2$,
$$E_2=\dfrac{Q_{2,in}}{4\pi r_2^2\epsilon_0 }=\dfrac{Q_{\rm inner}+Q_{\rm cavity}}{4\pi r_2^2\epsilon_0 }$$
Plug the known;
$$E_2=\dfrac{-4.524\times 10^{-9}+4.524\times 10^{-9}}{4\pi (0.08)^2(8.85\times 10^{-12})}$$
$$E_2=\color{red}{\bf 0}\;\rm N/C$$
At $r_3$,
$$E_3=\dfrac{Q_{1,in}}{4\pi r_3^2\epsilon_0 }=\dfrac{Q_{\rm cavity}+Q_{\rm inner}+Q_{\rm outer}}{4\pi r_3^2\epsilon_0 }$$
Plug the known;
$$E_3=\dfrac{(+4.524-4.524+12.57)\times 10^{-9}}{4\pi (0.12)^2(8.85\times 10^{-12})}$$
$$E_3=\color{red}{\bf 7.85\times 10^3}\;\rm N/C\tag {Outward}$$
The direction of this electric field is outward since the net inner charge of the whole surface is positive.
