What is the field inside a charged rubber sphere ( i.e a balloon) on whose surface , charges are uniformly distributed? In our physics textbook, we learned in applications of Gauss Law that the intensity of a field inside a hollow charged conducting sphere, is 0.
Given as flux inside such a sphere, $\phi$, is 0, so
$$
  \phi = \textbf{E.A}
$$
$$
  \phi \neq 0
$$
$$
  \textbf{A} \neq 0
$$
$$
  \therefore \textbf{E} = 0
$$
All the explanations I read online, (like this) only imply the case of the conducting sphere.
But there is an exercise question in the textbook, which i will quote verbatim so as to remove any ambiguity from my point of view.

Is $\textbf{E}$ necessarily zero inside a charged rubber balloon if the balloon is spherical? Assume that charge is distributed uniformly over the surface.

Now, I wasn't able to come up with an answer myself so I checked with a solution manual (Assume , that it is not credible), and I found it solved for the rubber sphere (balloon) the exact same way.
Was that right, is the field inside a rubber sphere , 0?
 A: Question specifically mentioned that the charge is distributed on the surface. So applying the Gauss law:
ϕ=EA=Q/ϵ
Select the Gaussian surface inside the sphere and apply the Gauss law. Then Q becomes zero and hence E is zero, when the charge distribution is symmetric over the surface.
A: For generality, assume the sphere is an insulator, has a radius R, and has a charge Q distributed uniformly throughout the sphere, what then is the electric field as a function of r, the radial distance?

As the comment by hsinghal says, the question stipulates that the charge is  on the outside, so then in the actual question,  the field inside, as with a conductor, would be zero.

Forget the stipulation for the moment that the charge is spread over the outside  of the sphere, just to find out how a charge inside an insulating body would behave.
If we compare the field emitting from a point charge Q to the electric field existing outside  the insulating sphere. Draw a Gaussian surface around the rubber/insulating shell.
Applying  Gauss's Law will produce, in both situations, 
$E = kQ/r^2$
So, outside a sphere of charge Q the field is identical to that from a point charge.
Now we have to find the electric field at the centre of the object.
What is the electric flux through a interior Gaussian sphere  r < R, and how much charge is enclosed by the sphere? 
The volume of the Gaussian sphere is:
$V_g = (4/3)\pi r^3$
The charge enclosed by the Gaussian surface is the charge per unit volume multiplied by the volume of the sphere. 
The uniform charge per unit volume $\rho $ in the insulating sphere is its total charge (Q) divided by its total volume.
$\rho =\frac  {Q}{(4/3)\pi R^3 }$
The charge enclosed by the gaussian surface is then:
$q_{enc} = Qr^3/R^3$
The flux through the gaussian surface is $EA = 4\pi r^2E$
Applying Gauss' Law:
Net flux = $\frac  {Q }{\epsilon_0}$

For the electric field inside the sphere we get:
$E =\frac   {Qr}{(4\pi \epsilon_0R^3)}$
So, internally  the field is proportional to $r$, and externally it's proportional to $1/r^2$. At the boundary, (the insulating shell), the two equations give the same value.
