Let $\mathcal{R}_n = \lim\limits_{k\to\infty} R(n,k)$. Your numbers on $\mathcal{R}_n$ doesn't feel right for $n \ge 3$.
For any fixed $n$ and large $k$, the centers of the small spheres of radius $r$ should be constrained to a sphere of radius $1-r$. To the small spheres, the space between two spherical shell of radius $1$ and $1-2r$ will look flat. So the optimal $(r,k)$ configuration will correspond to some sort of close packing of $S^{n-2}$ in a $\mathbb{R}^{n-1}$.
Let $\rho_{n}$ be the optimal packing density of $S^{n-1}$ in $\mathbb{R}^n$.
Let $\displaystyle\sigma_n = \frac{\pi^{n/2}}{\Gamma(\frac{n}{2}+1)}$ be the volume of the unit $n$-ball in $\mathbb{R}^n$. We know the "surface area' of the unit $n$-ball is given by $n\sigma_n$. This leads to
$$\begin{align}
& \sigma_{n-1} k\,r^{n-1} \approx n\sigma_n (1-r)^{n-1} \times \rho_{n-1}
\\
\implies & \mathcal{R}_n = \lim_{k\to\infty} k\,r^{n-1}(n,k) \approx n\frac{\sigma_n}{\sigma_{n-1}}\rho_{n-1} =
\frac{2\sqrt{\pi}\,\Gamma\left(\frac{n+1}{2}\right)}{\Gamma\left(\frac{n}{2}\right)}\rho_{n-1}
\end{align}$$
It is known that $\rho_1 = 1$, $\displaystyle \rho_2 = \frac{\pi}{2\sqrt{3}}$, $\displaystyle \rho_3 = \frac{\pi}{\sqrt{18}}\color{blue}{^{[1]}}$ and $\displaystyle \rho_8 \approx \frac{\pi^4}{384} ( 1 + O(10^{-14}) )\color{blue}{^{[2]}}$. This give us an estimate
$$\mathcal{R}_n \approx \begin{cases}
\pi,&n = 2\\
\\
\frac{2\pi}{\sqrt{3}},& n = 3\\
\\
\frac{\pi^2}{4\sqrt{2}}, & n = 4\\
\\
\frac{2\pi^4}{105},&n = 9
\end{cases}$$
Something very different from your guess when $n \ge 3$.
About what happens to $\mathcal{R}_n$ for large $n$, we know that
for $n \ge 115$, there is an upper bound for $\rho_n$ of the form$\color{blue}{^{[3]}}$:
$$\rho_n \le 2^{-(0.5990\ldots + o(1))n}$$
This means $\mathcal{R}_n$ converges to $0$ as $n$ tends to infinity.
Notes
$\color{blue}{[1]}$ T.C. Hales, A proof of the Kepler conjecture, Ann. of Math. (2) 162 (2005), 1065-1185. MR2179728 doi:10.4007/annals.2005.162.1065.
$\color{blue}{[2]}$ the packing density $\frac{\pi^4}{384}$ is achieved
by regular packing of spheres on an $E_8$ lattice in $\mathbb{R}^8$. The $O(10^{-14})$ error bound is given by H.Cohn and A. Kumar but I don't know the exact reference.
$\color{blue}{[3]}$ G.A.Kabatyanskii and V.I.Levenshetin. Bounds for packings on a sphere and in space (Russian), Problemy Peredaci Informacii 14 (1978), 3-25; English translation in Problems of Information Transmission 14 (1978), 1-17, MR0514023.
Best Answer
A simple approach for producing reasonable lower bounds is to use a face-centered cubic packing or a hexagonal packing (both have the optimal density, $\frac{\pi}{2\sqrt{3}}\approx 74\%$, in the unconstrained space) and to count the number of spheres met by $x^2+y^2+z^2=(20)^2$. Recalling that the optimal packing density in the plane is $\frac{\pi\sqrt{3}}{6}$, in a sphere with radius $20$ it should be possible to pack around
$$ \frac{\pi}{2\sqrt{3}}\cdot 20^3 - \frac{\pi\sqrt{3}}{6}\cdot 4(20)^2 \approx\color{red}{5804}$$ spheres, but not many more. The estimated density is so $\approx 72.5\%$.