Metallic Spheres

We can use the fact that this system can be considered to be two spherical capacitors connected in parallel (1,2 and 2,3), since they are at same potential.

We start with putting charges on outer and inner surfaces of spheres using the fact that net charge within a conductor is zero. Let outer surface of innermost conductor have charge $q$ . Then using the above fact we find subsequent charges are $-q,q',-q',q''$ $($ From inside to outside $)$ . Now applying the fact that inner and outer spheres are grounded $($ i.e. their surface potential is zero $)$ we have the following equations:

$\displaystyle\frac{q}{10}+\displaystyle\frac{q'-q}{30}+\displaystyle\frac{q''-q'}{40}=0\;\;\;---(i)$

$\displaystyle\frac{q''-q'}{40}+\displaystyle\frac{q'}{40}=0\;\;\;---(ii)$

From $(ii)$ we have $q"=0$ , which we could also have got from the fact that charge inside a conductor is zero. Plugging this into $(i)$ we have $q=\displaystyle\frac{-q'}{8}$ .

Now charge on middle sphere is $Q=q'-q=\displaystyle\frac{9q'}{8}$ and it's potential $V=\displaystyle\frac{k\times\,q'}{30\times10^{-2}}-\displaystyle\frac{k\times\,q'}{40\times10^{-2}}=\displaystyle\frac{10\times\,k\times\,q'}{12}$ .

So $\displaystyle\frac{Q}{V}=C_{self}=\boxed{1.5\times10^{-10}}.$