Electric field $\propto \frac{1}{r^3}$ 
If some charge is given to a solid metallic sphere, the field inside remains zero and by Gauss’s law all the charge resides on the surface. Suppose now that Coulomb’s force between two charges varies as $\frac{1}{r^3}$. Then, for a charged solid metallic sphere 
A) field inside will be zero and the charge density inside will be zero.
B) field inside will not be zero and the charge density inside will not be zero. 
C) field inside will not be zero and the charge density inside will be zero.
D) field inside will be zero and the charge density inside will not be zero.

Considering the case for a single point charge,
$$ \int \vec{E} \cdot \vec{ds} \propto \frac{Q}{r}.$$
The electric flux is dependent on $r$ so Gauss' law is not applicable. On considering the case of electrostatic equilibrium the net electric field inside the charged body should be zero (because if it isn't then the charges must experience some net force and hence not be in equilibrium), but I'm unable to conclude anything about the charge distribution from this point onwards.
 A: Although the Coulomb force law changes to $\propto 1/r^3$, it is still repulsive between like charges. Thus, the charges would still want to stay away from each other as much as possible. The resulting charge distribution should be spherically symmetric.
Assume that, intuitively, the charge distributes uniformly on the boundary of the sphere. In the real world, where Coulomb force law is $\propto 1/r^2$, the electric field in the interior of the sphere is exactly zero. This follows readily from Gauss' law. However, I would like to present an alternative argument for why the field should be zero, which we can adapt to the $1/r^3$ case.

Consider the (cross-section of the) charged sphere above. We want to know the field at an arbitrary interior point (drawn in red). There is no loss of generality as it is always possible to cut the sphere into halves while passing through two specific points. Look at the two yellow charged patches, they are both constructed to subtend a solid angle $\Omega$. The distances to the field point are labelled as $r_1$ and $r_2$. When $\Omega \to 0$, the ratio of the area of the patches approaches $r_1^2/r_2^2$. (Recall that $dS = r^2 \sin \theta d \theta d\phi$, and both patches are slanted in the same way.) This means that patch 1 carries more charge. If the Coulomb force is $\propto 1/r^2$, then the Coulomb force due to the two patches will be equal in magnitude, and opposite in direction, and hence cancel. All opposite patches will cancel in this way, and the net electric field is zero.
Now we go to the world with Coulomb fore $\propto 1/r^3$. This means that charges further away will contribute less electric field. As a result, if you place a test charge at the red spot, it will experience a weaker electric force from the lower "hemisphere". The test charge will be repelled towards the center of the sphere. If you started with a spherical shell of charge, it is in a high-energy configuration. If you allow the charges to relax, they will start falling into the interior of the sphere. The equilibrium charge distribution will therefore be non-zero in the interior.
As you correctly pointed out, the charge distribution will be such that the resulting interior field is zero. Due to spherical symmetry, we can write the charge density as $\rho(r)$. Consider some charges at $r=a$, at equilibrium, they must experience no net force. This means that the force from those charges with $r<a$ balances the force from the charges $r>a$. To calculate the force in this way, you need to know the force due to an infinitesimal thin shell of charge density $\sigma = \rho dr$, and then integrate the force over all $0<r<R$. There are two integrations involved, and I haven't tried either. But at least I know that $\rho$ is non-zero. If you manage to do it please update the answer.
