Vector field ${\bf F}$ with $\int_S {\bf F}\cdot{\bf n}\ dS=c$ Find a vector 
field ${\bf F}$ on $
{\bf R}^3$ with $$\int_S {\bf F}\cdot{\bf n}\ dS=c > 0 \tag{1} $$ where $S$ is any closed surface containing $0$ and ${\bf n}$ is 
normal 
Here there is a solution $\frac{k}{r^3} (x,y,z)$. Note that from divergence therem we know that ${\bf F}$ has
divergence $0$ so that $(x,y,-2z)$ is possible. But this solution does not satisfy $(1)$. 
So the solution is unique ?

When they talk about  "closed surfaces $S$ containing $0$" they tacitly mean that such $S$ should bound a compact body $B\subset{\mathbb R}^3$ which contains $0$ in its interior. Now we cannot have arbitrarily tiny such surfaces giving a fixed value $c>0$ for the integral in question unless something terrible happens at $0$. 
You have remarked that the flow field
$${\bf G}(x,y,z):=\left({x\over r^3},{y\over r^3},{z\over r^3}\right)$$
could play a rôle in this question, and you are right: This field is divergence free outside of $0$ and has flow integral
$$\int_{S_R}{\bf G}\cdot {\bf n}\>{\rm d}\omega=4\pi$$
when integrated over the sphere $S_R$ of radius $R$ centered at $0$. Using Gauss' theorem, applied to $B\setminus S_R$ with $R\ll1$ it follows that the field
$${\bf F}:={c\over4\pi}{\bf G}\tag{1}$$
has flow $c$ through any surface of the kind described in the first paragraph, whence is a solution of the problem.
But this ${\bf F}$ is not the only solution of the problem at hand: Add to ${\bf F}$ an arbitrary $C^1$-vector field ${\bf v}$ which has divergence $0$ in all of ${\mathbb R}^3$. Then ${\bf F}+{\bf v}$ again solves the problem; and it is not difficult to show that all solutions are obtained in this way.
