= Solution
For static fields, the <Maxwell equations> give $\nabla\times B=\mu_0J$ and $\nabla\cdot B=0$. Write $B=\nabla\times A$. A <gauge transformation> $A\mapsto A+\nabla\chi$ leaves $B$ unchanged; choose $\Delta\chi=-\nabla\cdot A$ to impose the <Coulomb gauge>. Then
$$
\nabla\times(\nabla\times A)=\nabla(\nabla\cdot A)-\Delta A=\mu_0J,
\qquad \boxed{-\Delta A=\mu_0J.}
$$
For a localized conserved current with the potential chosen to vanish at infinity, the supplied fundamental solution gives the <magnetic vector potential>
$$
A(x)=\frac{\mu_0}{4\pi}\int\frac{J(r)}{|x-r|}\,d^3r.
$$
Its divergence vanishes by integration by parts and $\nabla\cdot J=0$. For a filamentary loop the current integral becomes $I\oint_Ldr$, so
$$
\boxed{A(x)=\frac{\mu_0I}{4\pi}\oint_L\frac{dr}{|x-r|}.}
$$
For $R=|x|$ much larger than the fixed loop size, expand $|x-r|^{-1}=R^{-1}+(x\cdot r)R^{-3}+O(R^{-3})$. The leading constant term integrates to zero because $\oint dr=0$. Therefore
$$
A(x)=\frac{\mu_0I}{4\pi R^3}\oint(x\cdot r)\,dr+O(R^{-3}).
$$
The vector triple-product identity gives $x\times(r\times dr)=r(x\cdot dr)-(x\cdot r)dr$. Integrating the derivative of $(x\cdot r)r$ around the closed loop shows $\oint r(x\cdot dr)=-\oint(x\cdot r)dr$. Consequently
$$
\boxed{A(x)=-\frac{\mu_0I}{4\pi R^3}\oint\frac12x\times(r\times dr)+O(R^{-3}).}
$$
Equivalently, $A=\mu_0m\times x/(4\pi R^3)+O(R^{-3})$, where $m=(I/2)\oint r\times dr$ is the <magnetic dipole moment>. The leading displayed potential is of order $R^{-2}$; the remainder is smaller by a factor $R^{-1}$.
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