= Solution
Write $r=|x|$, $u(x)=\bar u(r)$, and let $g(r)$ be the <radial profile> of the <radial derivative> $\nabla u(x)\cdot x/|x|$, which may be assumed to be a <radial function> here. The <spherical coordinate system> gives
$$
\int_A|u|^6\,dx=4\pi\int_1^2|\bar u(r)|^6r^2\,dr,\qquad
4\pi\int_1^2|g(r)|^6r^2\,dr
\leq\int_A|\nabla u|^6\,dx.
$$
Since $1\leq r^2\leq4$, both $\bar u$ and $g$ belong to the <Lp space> $L^6(1,2)$. It remains to identify the <weak derivative> of the <radial profile> directly.
For $\phi\in C_c^\infty(1,2)$, each component of $\phi(r)x/r$ is a <test function> on $A$. Sum the three defining <weak derivative> identities, using the stated <divergence> formula:
$$
\begin{aligned}
4\pi\int_1^2g(r)\phi(r)r^2\,dr
&=\int_A\nabla u\cdot\frac{x}{r}\phi(r)\,dx\\
&=-\int_Au\left(\phi'(r)+\frac2r\phi(r)\right)dx\\
&=-4\pi\int_1^2\bar u(r)\bigl(r^2\phi(r)\bigr)'\,dr.
\end{aligned}
$$
Given an arbitrary <test function> $\zeta\in C_c^\infty(1,2)$, choose $\phi(r)=\zeta(r)/r^2$. Dividing by $4\pi$ gives $\int g\zeta=-\int\bar u\zeta'$, exactly the definition of the <weak derivative>. We have proved
$$
\boxed{\bar u\in W^{1,6}(1,2),\qquad\bar u'=g\ \text{almost everywhere},\qquad
\|\bar u\|_{W^{1,6}(1,2)}\leq C\|u\|_{W^{1,6}(A)}.}
$$
This <radial Sobolev profile on an annulus> estimate follows from the two weighted <integrals> above and equivalence of the usual <Sobolev norms>.
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