= Solution
Put $\mathbf R=\mathbf r-\mathbf r'$. The <magnetic field> is the <curl> of the <magnetic vector potential>. Since $\mathbf J(\mathbf r')$ does not depend on the differentiated variable $\mathbf r$,
$$
\nabla_{\mathbf r}\times\frac{\mathbf J(\mathbf r')}{R}
=\nabla_{\mathbf r}(R^{-1})\times\mathbf J(\mathbf r')
=-\frac{\mathbf R}{R^3}\times\mathbf J(\mathbf r')
=\frac{\mathbf J(\mathbf r')\times\mathbf R}{R^3}.
$$
Taking the <curl> under the source <integral> consequently gives the <Biot-Savart law>,
$$
\mathbf B(\mathbf r)=\frac{\mu_0}{4\pi}\int\frac{\mathbf J(\mathbf r')\times(\mathbf r-\mathbf r')}{|\mathbf r-\mathbf r'|^3}\,d^3r'.
$$
Outside the compact source this differentiation is ordinary differentiation of a smooth kernel. For regular volume currents, the locally integrable $R^{-2}$ derivative kernel gives the same formula at interior points.
For the loop, choose positive current circulating counterclockwise as viewed from the positive $z$-axis. Parametrize its source position and directed line element by
$$
\mathbf r'=a(\cos\varphi,\sin\varphi,0),\qquad d\boldsymbol\ell'=a(-\sin\varphi,\cos\varphi,0)\,d\varphi.
$$
At $(0,0,z)$ the denominator is the constant $(a^2+z^2)^{3/2}$, while
$$
d\boldsymbol\ell'\times\mathbf R=(az\cos\varphi,az\sin\varphi,a^2)\,d\varphi.
$$
The transverse terms integrate to zero around the loop. The <Biot-Savart law> therefore yields
$$
\boxed{\mathbf B(0,0,z)=\frac{\mu_0Ia^2}{2(a^2+z^2)^{3/2}}\,\mathbf e_z.}
$$
The <magnetic field> is axial, with the stated magnitude for $I>0$; reversing the <electric current> reverses its direction.
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