= Solution
Choose an origin near the bounded loop and write $\mathbf r'$ for the source position. The <magnetic vector potential> and its far-field <multipole expansion> are
$$
\mathbf A(\mathbf r)=\frac{\mu_0 I}{4\pi}\oint_C\frac{d\mathbf r'}{|\mathbf r-\mathbf r'|},\qquad
\frac1{|\mathbf r-\mathbf r'|}=\frac1r+\frac{\mathbf r\cdot\mathbf r'}{r^3}+O(r^{-3}),
$$
with the loop held fixed as $r=|\mathbf r|\to\infty$. The first term integrates to zero since $C$ is closed. Define its oriented <vector area> by
$$
\mathbf S=\frac12\oint_C\mathbf r'\times d\mathbf r'=\int_\Sigma\mathbf n\,dS.
$$
The second equality is <Stokes theorem>, applied componentwise to any oriented spanning surface. In particular the <vector area> is independent of the choice of spanning surface. From $\oint_C d(r'_i r'_j)=0$, the matrix $\oint_C r'_i\,dr'_j$ is antisymmetric, which gives $\oint_C(\mathbf r\cdot\mathbf r')d\mathbf r'=\mathbf S\times\mathbf r$. Hence
$$
\boxed{\mathbf m=I\mathbf S,\qquad \mathbf A(\mathbf r)=\frac{\mu_0}{4\pi}\frac{\mathbf m\times\mathbf r}{r^3}+O(r^{-3})}.
$$
The orientation follows the current by the <right-hand rule>. For $r\ne0$, taking the <curl> of this leading <magnetic vector potential> gives the <magnetic dipole field>
$$
\boxed{\mathbf B(\mathbf r)=\frac{\mu_0}{4\pi}\left(\frac{3(\mathbf m\cdot\mathbf r)\mathbf r}{r^5}-\frac{\mathbf m}{r^3}\right)+O(r^{-4})}.
$$
For a planar loop, $\mathbf S$ is the signed area times the unit <normal vector>. For a nonplanar loop it is the oriented <vector area>, rather than the scalar area of an arbitrarily chosen surface.
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