Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The easiest way to solve this problem is to use the Gaussian surfaces.
Assume that there is a Gaussian sphere inside the sphere of radius $R_2$ where $R_1\lt r\lt R_2$. See the dashed circle in the figure below that represents the Gaussian sphere.
Now let's assume that $R_2-R_1\lt\lt R_1$ which means that the inner sphere is very close to the outer shell sphere.
So we can consider this system as a parallel plate capacitor.
We know, from Gauss’s Law for the Gaussian surface, that
$$\oint EdA=\dfrac{Q_{\rm enclosed}}{\epsilon_0}$$
Since the system is considered as a parallel plate capacitor, the electric field is constant inside it and since the symmetry of the system, the electric field is always normal to the surface.
Thus,
$$E\oint dA=\dfrac{Q_{\rm enclosed}}{\epsilon_0}$$
$$E(4\pi r^2)=\dfrac{Q_{\rm enclosed}}{\epsilon_0}$$
Solving for $E$,
$$E=\dfrac{Q_{\rm enclosed}}{ 4\pi\epsilon_0 r^2 }$$
It is obvious that the enclosed charge here is the charge in the inner sphere which is considered as the inner plate, so the enclosed charge is equal to the charge of the capacitor; $Q_{\rm enclosed}=Q$
$$E=\dfrac{Q}{ 4\pi\epsilon_0 r^2 }\tag 1$$
Now we need to find $\Delta V_C$ where we know that $\Delta V=-\int Edr $
$$\Delta V_C=-\int_{R_1}^{R_2}Edr$$
Plug from (1),
$$\Delta V_C=-\dfrac{Q}{ 4\pi\epsilon_0 }\int_{R_1}^{R_2} \dfrac{dr}{r^2}$$
$$\Delta V_C=-\dfrac{Q}{ 4\pi\epsilon_0 }(-1) \dfrac{1}{r}\bigg|_{R_1}^{R_2}$$
$$\Delta V_C= \dfrac{Q}{ 4\pi\epsilon_0 } \left[ \dfrac{1}{R_1}-\dfrac{1}{R_2}\right]\tag 2$$
Recalling that $\Delta V_C=Q/C$
So,
$$C=\dfrac{Q}{\Delta V_C}$$
Plug from (2),
$$C=\dfrac{Q}{ \dfrac{Q}{ 4\pi\epsilon_0 } \left[ \dfrac{1}{R_1}-\dfrac{1}{R_2}\right]}$$
$$\boxed{C=\dfrac{4\pi\epsilon_0 }{ \left[ \dfrac{1}{R_1}-\dfrac{1}{R_2}\right]}}$$
$$\color{blue}{\bf [b]}$$
Let's assume that the gap is $d$, so $R_2=R_1+d$.
Plug that into the boxed formula above,
$$ C=\dfrac{4\pi\epsilon_0 }{ \left[ \dfrac{1}{R_1}-\dfrac{1}{R_1+d}\right]} $$
Solving for $R_1$,
$$ \dfrac{1}{R_1}-\dfrac{1}{R_1+d} =\dfrac{4\pi\epsilon_0 }{ C} $$
$$ \dfrac{R_1+d-R_1}{R_1^2+R_1d} =\dfrac{4\pi\epsilon_0 }{ C} $$
$$ \dfrac{d}{R_1^2+R_1d} =\dfrac{4\pi\epsilon_0 }{ C} $$
$$ R_1^2+R_1d -\dfrac{Cd}{4\pi\epsilon_0}=0$$
Plug the known;
$$ R_1^2+(1\times 10^{-3})R_1 - (100\times 10^{-12})(1\times 10^{-3})(9\times 10^9) =0$$
So whether $R_1 =\bf -0.0305 \;\rm m$ or $R_1=\bf 0.0295$
Therefore,
$$R_1=\bf 2.95\;\rm cm$$
and hence,
$$R_2=2.95+0.1=\bf 3.05\;\rm cm$$
So their diameters are
$$D_1=\color{red}{\bf 5.9}\;\rm cm$$
$$D_2=\color{red}{\bf 6.1}\;\rm cm$$
