Part A)
First of all, let's convert the radii of the inner and the outer sphere:


The capacitance of a spherical capacitor which consist of two shells with radius rA and rB is


Then, from the usual relationship between capacitance and voltage, we can find the charge Q on each sphere of the capacitor:

Now, we can find the electric field at any point r located between the two spheres, by using Gauss theorem:

from which

In part A of the problem, we want to find the electric field at r=11.1 cm=0.111 m. Substituting this number into the previous formula, we get

And so, the energy density at r=0.111 m is

Part B) The solution of this part is the same as part A), since we already know the charge of the capacitor:

. We just need to calculate the electric field E at a different value of r: r=16.4 cm=0.164 m, so

And therefore, the energy density at this distance from the center is