I have solved it this way:
Being a sphere, the normal vector is the radial vector, hence we have $d\underline{S} = \underline{\hat{r}}dA$. Therefore we have $$\iint_S\frac{\underline{r}}{r^3}\cdot\underline{\hat{r}}dA = \iint_S\frac{1}{r^2}dA = \frac{1}{a^2}\iint_SdA = \frac{4\pi a^2}{a^2} = 4\pi$$ since the surface area of a sphere is $4\pi a^2$.
However, I thought I could solve it through the integral theorems as well, my idea was that the surface of the sphere is a closed surface containing the volume of the sphere, so I could use the divergence theorem. So I calculate $$\nabla\cdot\left(\frac{\underline{r}}{r^3}\right)=\nabla\left(\frac{1}{r^3}\right)\cdot \underline{r}+\frac{1}{r^3}(\nabla\cdot\underline{r}) =-\frac{3}{r^4}\underline{\hat{r}}\cdot \underline{r}+\frac{3}{r^3}=0$$ and hence by divergence theorem I get $$\iint_S\underline{F}\cdot d\underline{S} = \unicode{x222F}_{\partial V}\underline{F}\cdot d\underline{S} = \iiint_V0dV = 0$$ Why is the result different? I suppose that both the volume and the surface which is the boundary of the volume satisfy the assumptions of the Divergence theorem,so either this is wrong or the above one is. Can you help me?