Electrostatics - Homework: Relative permitivitty of spherical capacitor
I'm not going to give you the answer, but I'll give you a hint and hopefully you can work it out from there. Disclaimer: It has been about 3 years since I studied this topic, so you should check that everything I am writing makes sense based on what you have been taught - if it is wrong, don't come crying to me if you get a bad mark it is your responsibility to verify and understand everything.

You have the following two equations:

$$E_1^{(1)}=\frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2},a&lt;r&lt;b$$

$$E_2^{(1)}=\frac{Q}{4\pi\epsilon_0\epsilon_{r2}r^2},b&lt;r&lt;c$$ 

You know that when dielectric 2 is removed, the following is true:

$$E_1^{(2)}=\frac{2}{3}\times E_1^{(1)}$$

In order for the above to be true, the only thing that can change is the charge \$Q\$, so based on this we can say:

$$\frac{Q^{(2)}}{4\pi\epsilon_0\epsilon_{r1}r^2} = \frac{2}{3}\frac{Q^{(1)}}{4\pi\epsilon_0\epsilon_{r1}r^2}$$ 

You can find \$Q^{(2)}\$ is in terms of \$Q^{(1)}\$ such that the above is true.

We also know from the question that 

$$E_2^{(2)}=2\times E_2^{(1)}$$

We can also say that:

$$\frac{Q^{(2)}}{4\pi\epsilon_0 r^2} = 2\frac{Q^{(1)}}{4\pi\epsilon_0\epsilon_{r2} r^2}$$ 

Now if you substitute for \$Q^{(2)}\$ you get an equation for which everything is the same on both sides, except one side has an \$\epsilon_{r2}\$ and the other side has a constant scale factor. Rearrange and you find a value for \$\epsilon_{r2}\$.



Next we need to know what \$\epsilon_{r1}\$ is. 

You have worked out that (I haven't checked if this is correct, but you put it in your question so I am going with what you have worked out):

$$U^{(1)}=\int_a^b \frac{Q}{4\pi\epsilon_0\epsilon_{r1}r^2}\mathrm dr+ \int_b^c \frac{Q}{4\pi\epsilon_0\epsilon_{r2}r^2}\mathrm dr=\frac{Q^{(1)}}{4\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{\epsilon_{r2}bc}\right)$$

We can then say that (note that there is no need to do the integration again, we just say that \$Q\$ and \$\epsilon_{r2}\$ have changed as shown):

$$U^{(2)}=\frac{Q^{(2)}}{4\pi\epsilon_0}\left(\frac{b-a}{\epsilon_{r1}ab}+\frac{c-b}{bc}\right)$$

You have two equations for \$U\$ now. Now you know that this voltage is constant (as per the question), so that means in both cases \$U\$ must be the same, so you can equate your two equations.

Now do some simplification by first substituting the relationship between \$Q^{(2)}\$ and \$Q^{(1)}\$, and also substituting \$\epsilon_{r2}\$. Once that all simplifies down you will get a value for \$\epsilon_{r1}\$