= Solution
{c}
Rearrange the source's condensate-current expression and integrate around a closed path lying in the superconducting material:
$$
\hbar\oint\nabla\theta\cdot d\mathbf l=q\oint\mathbf A\cdot d\mathbf l+\frac{m}{n_sq}\oint\mathbf J_s\cdot d\mathbf l.
$$
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>:
$$
\boxed{\Phi+\frac{m}{n_sq^2}\oint\mathbf J_s\cdot d\mathbf l=\frac{nh}{q}.}
$$
Reversing the winding orientation absorbs the sign of $q$. Since a <Cooper pair> has $|q|=2e$, the positive <superconducting flux quantum> is
$$
\boxed{\Phi_0=\frac{h}{2e}\simeq2.068\times10^{-15}\,\mathrm{Wb}.}
$$
For 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.
\b[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 $\Phi=L_gI=\Phi_0/2$. This concrete counterexample satisfies the given current law while its magnetic flux is not an integer multiple of $\Phi_0$.
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