= Solution
The magnetostatic <Ampère-Maxwell equation> is $\nabla\times B=\mu_0J$. Substituting $B=\nabla\times A$ and using the <Coulomb gauge> $\nabla\mathbin{\cdot}A=0$ gives
$$
\mu_0J=\nabla\times(\nabla\times A)
=\nabla(\nabla\mathbin{\cdot}A)-\nabla^2A
=-\nabla^2A.
$$
The free-space <Green function of the Laplacian> therefore gives
$$
A(x)=\frac{\mu_0}{4\pi}
\int_{\mathbb R^3}\frac{J(x')}{|x-x'|}\,d^3x'.
$$
For the thin wire current stated in the question this becomes
$$
\boxed{
A(x)=\frac{\mu_0I}{4\pi}\oint_C\frac{dx'}{|x-x'|}}.
$$
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