 Solution

Rearrange the source's condensate-current expression and integrate around a closed path lying in the superconducting material:Single-valuedness of the superconducting wavefunction requires $\oint \nabla \theta \cdot d\mathbf{l}=2\pi n$, and Stokes theorem gives $\oint \mathbf{A}\cdot d\mathbf{l}=\Phi$. Thus the exact result is fluxoid and flux quantization in a superconductor:Reversing the winding orientation absorbs the sign of $q$. Since a Cooper pair has $|q|=2e$, the positive superconducting flux quantum isFor a thick ring, choose the path deep enough that the supercurrent is negligible; then the enclosed magnetic flux is $n{\Phi }_0$. This is the flux quantization usually intended in the question.

Qualification of the unqualified flux claim: a thin ring can have a nonzero kinetic contribution around every available path, so it is the fluxoid that is exactly quantized. For example, with zero applied flux and geometric and kinetic inductances $L_g=L_k$, a metastable winding-$1$ state has $(L_g+L_k)I={\Phi }_0$ but $\Phi =L_{g}I={\Phi }_0/2$. This concrete counterexample satisfies the given current law while its magnetic flux is not an integer multiple of ${\Phi }_0$.