= Solution
A positive matching root $x_*$ gives $s=x_*^2/R^2$. Substituting the root value requested by the question gives the intended expression
$$
\boxed{s_{\rm intended}=\frac{\pi^2}{R^2}.}
$$
After the index correction in part (iii), this is indeed the lowest mode in the $\ell=1$ sector, because its condition is $j_0(x)=\sin x/x=0$ and its first positive root is $\pi$. It is also the minimum over $\ell\geq1$.
However, \b[the literal scalar problem includes $\ell=0$], for which matching gives $\cos x=0$. Its first positive root is $\pi/2$, so the actual minimum for the written scalar equations and stated boundary conditions is
$$
\boxed{s_{\rm scalar,min}=\frac{\pi^2}{4R^2}.}
$$
An explicit counterexample to the larger proposed minimum is $B_{\rm in}=j_0(\pi r/(2R))$, $B_{\rm out}=2R/(\pi r)$, with no angular dependence. At $R$, both values are $2/\pi$ and both radial derivatives are $-2/(\pi R)$.
To verify minimality, put $u=rF_\ell$. The radial problem is $-u''+\ell(\ell+1)u/r^2=su$, $u(0)=0$, $u'(R)+\ell u(R)/R=0$. Its <Rayleigh quotient> is
$$
s=\frac{\int_0^R[|u'|^2+\ell(\ell+1)|u|^2/r^2]dr+\ell|u(R)|^2/R}{\int_0^R|u|^2dr}.
$$
For $\ell=0$ this is the Dirichlet-Neumann interval problem, whose lowest eigenvalue is $\pi^2/(4R^2)$. The additional nonnegative terms increase with $\ell$, proving the scalar minimum and the minimum-at-$\ell=1$ statement for the restricted family. A monopole-free magnetic multipole interpretation would exclude $\ell=0$ and yield the intended value, but that extra restriction is not supplied by the stated scalar model. The claim about the first zero is not valid for every $f_\ell$ as printed.
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