$\quad F \propto \frac{1}{r^{3}}$
$\Rightarrow \text { Electric field, } E \propto \frac{1}{r^{3}} \Rightarrow E=\frac{k q}{r^{3}}$
Now, for Gaussian surface $(r < R)$.
Electric flux linked with surface is
$\phi=\int \frac{k q}{r^{3}} \cdot 2 \pi r d r$
$=2 \pi k q \int \frac{d r}{r^{2}}=2 \pi k q\left(-\frac{1}{r}\right)$
As flux is non-zero, charge density is also non-zero. Also, by symmetry of charge distribution electric field is zero.
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