= Solution
Set $r=\ell_0\widetilde r$, $R=\sqrt{\mu/U}\,f(\widetilde r)$, using the <healing length> $\ell_0=\hbar/\sqrt{2m\mu}$ and $U,\mu>0$. For zero external potential, division by $\mu R$ produces
$$
\boxed{f''+\frac1{\widetilde r}f'-\frac{\mathcal N^2}{\widetilde r^2}f+(1-f^2)f=0,
\qquad f(\infty)=1.}
$$
Primes now denote dimensionless-radius derivatives. Let $f=1-a/\widetilde r^2+O(\widetilde r^{-4})$. Then $f''+f'/\widetilde r=O(\widetilde r^{-4})$, while the centrifugal term is $-\mathcal N^2/\widetilde r^2+O(\widetilde r^{-4})$ and $(1-f^2)f=2a/\widetilde r^2+O(\widetilde r^{-4})$. Their leading balance gives $2a=\mathcal N^2$. Therefore the <inverse-square tail of a Gross–Pitaevskii vortex> yields
$$
\boxed{\lim_{\widetilde r\to\infty}\widetilde r^2[1-f(\widetilde r)]=\frac{\mathcal N^2}{2}.}
$$
For a singly quantized vortex the answer is $1/2$. This is the amplitude deficit; the density deficit $1-f^2$ has limiting coefficient $\mathcal N^2$ instead.
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