= Solution
For this question use the normalized <magnetic scalar potential> $\mathcal U$ defined by $\mathbf B=(\mu_0/4\pi)\nabla\mathcal U$. The previous question instead includes $\mu_0/4\pi$ in its scalar potential; its $U$ equals $(\mu_0/4\pi)\mathcal U$. This normalization change in the paper must be kept separate from the kernel calculation.
Take the radial component of the <Geselowitz formula>. On the spherical boundary, $\mathbf n'$ is parallel to $\mathbf r'$, and
$$
[\mathbf n'\times(\mathbf r-\mathbf r')]\cdot\widehat{\mathbf r}=0.
$$
Thus the surface-current term contributes no radial <magnetic field>. The primary <Biot-Savart law> term gives, along a fixed direction $\widehat{\mathbf r}$,
$$
\partial_r\mathcal U
=\left[\mathbf Q\times\frac{\mathbf r-\mathbf r_0}{|\mathbf r-\mathbf r_0|^3}\right]\cdot\widehat{\mathbf r}
=-\frac{(\mathbf Q\times\mathbf r_0)\cdot\widehat{\mathbf r}}
{|r\widehat{\mathbf r}-\mathbf r_0|^3}.
$$
Choose $\mathcal U\to0$ at infinity. Integrating this equation fixes the full potential along every ray, giving
$$
\mathcal U=(\mathbf Q\times\mathbf r_0)\cdot\widehat{\mathbf r}
\int_r^\infty\frac{ds}{|s\widehat{\mathbf r}-\mathbf r_0|^3}.
$$
To express it as a source-position derivative, define
$$
\phi(\mathbf r;\mathbf r_0)=\int_r^\infty\frac{ds}{s|s\widehat{\mathbf r}-\mathbf r_0|}.
$$
With $\mathbf W=\mathbf Q\times\mathbf r_0$, differentiating the integrand gives
$$
\mathbf W\cdot\nabla_{\mathbf r_0}\frac1{|s\widehat{\mathbf r}-\mathbf r_0|}
=\frac{s\mathbf W\cdot\widehat{\mathbf r}}{|s\widehat{\mathbf r}-\mathbf r_0|^3},
$$
so $\mathcal U=\mathbf W\cdot\nabla_{\mathbf r_0}\phi$ as required.
Let $b=|\mathbf r_0|$ and $P=|\mathbf r-\mathbf r_0|$. An elementary antiderivative gives the <spherical current-dipole logarithmic kernel>
$$
\boxed{\phi(\mathbf r;\mathbf r_0)=\frac1b\log\frac{r+P+b}{r+P-b}.}
$$
For $r>a>b$, both logarithm arguments are positive. Its radial derivative is $-1/(rP)$ and its limit at infinity is zero, which verifies the preceding integral. At $b=0$ use the continuous limit $\phi=1/r$.
For a direct check, differentiating along $\mathbf W$ leaves $b$ unchanged and gives $\mathbf W\cdot\nabla_{\mathbf r_0}P=-\mathbf W\cdot\mathbf r/P$. Since
$$
F=rP^2+P\mathbf r\cdot(\mathbf r-\mathbf r_0)
=\frac P2[(r+P)^2-b^2],
$$
one obtains
$$
\boxed{\mathcal U=\frac{(\mathbf Q\times\mathbf r_0)\cdot\mathbf r}{F}.}
$$
Its gradient is the <Sarvas formula>. The sphere radius and <electrical conductivity> disappear from this exterior magnetic result because spherical symmetry removed the radial contribution of the return currents. The original derivation and spherical-conductor assumptions are discussed in https://pubmed.ncbi.nlm.nih.gov/3823129/[Sarvas's paper].
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