= Solution
The <determinant> condition is $r'(r/R)^2=1$, or $r^2r'=R^2$. Integration gives $r^3=R^3+C$ and the outer boundary value fixes $C=t^3-1$. Thus
$$
\boxed{r(R)=(R^3+t^3-1)^{1/3}.}
$$
For $t>1$, $r(R)$ tends to the positive cavity radius $(t^3-1)^{1/3}$ as $R\downarrow0$. Its tangential stretches grow like that radius divided by $R$, so $|F|\to\infty$. The <deformation> is singular at the origin and is interpreted as a finite-energy Sobolev <deformation>, with a point value irrelevant to its energy.
Since $r'=R^2/r^2$ and the <determinant> penalty is zero almost everywhere,
$$
\boxed{E_{\mathrm{cav}}(t)=4\pi\int_0^1\left(\frac{R^6}{r(R)^4}+2r(R)^2\right)dR.}
$$
For fixed $t>1$, the first integrand is $O(R^6)$ near zero and the second is bounded. Thus the energy is finite despite the unbounded <gradient>. In particular the <gradient> is square integrable, as required for this competitor.
For large $t$, $r(R)/t\to1$ uniformly on $[0,1]$. The first <integral> is $O(t^{-4})$, while $\int_0^1r(R)^2dR\sim t^2$. Consequently
$$
\boxed{E_{\mathrm{cav}}(t)\sim8\pi t^2.}
$$
The spherical <volume> factor $R^2dR$ is what makes the $R^{-1}$ tangential-<gradient> singularity integrable in three dimensions.
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