One constraint is the recommendation $F=\frac{d^2}{b\lambda}\geq 1$, in this case with $d=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, and $b$ the distance between circular obstacle and observer.
The distance between Earth and the observer should hence be
$b\leq \frac{d^2}{\lambda}=\frac{(12700\cdot10^3\mbox{ m})^2}{600\cdot 10^{-9}\mbox{ m}}=481.67\cdot 10^{18}\mbox{ m}$
Another constraint is the surface roughness of the circular object:
$\Delta r < \sqrt{r^2 + \lambda\frac{gb}{g+b}}-r$, with $r=6350\mbox{ km}$ the radius of the circular obstacle (here Earth), $g$ the distance between the point light source and the circular obstacle, and $b$ the distance between the circular obstacle and the screen.
To simplify calculations, say $g\gg b$. Then approximately
$\Delta r < \sqrt{r^2 + \lambda\frac{gb}{g}}-r = \sqrt{r^2 + \lambda b}-r$.
After adding $r$ and squaring you get
$(\Delta r +r)^2<r^2+\lambda b$.
This simplifies to
$(\Delta r)^2+2r\Delta r<\lambda b$.
Assume $\Delta r\ll r$, and neglect the second order $(\Delta r)^2$ to get
$2r\Delta r<\lambda b$. Divide by $\lambda$ to get an approximate constraint for $b$ as
$b>\frac{2r\Delta r}{\lambda}$.
With $2r=12700 \mbox{ km}$ about the diameter of Earth, $\lambda=600 \mbox{ nm}$ some wavelength of visible light, we get
$b>\frac{12700\cdot 10^3\mbox{ m}\cdot\Delta r}{600\cdot 10^{-9}\mbox{ m}}
= 21.1667\cdot 10^{12}\Delta r$.
The two constraints allow for reasonable values of $\Delta r$.
Assume a surface roughness of Earth of e.g. $\Delta r = 1\mbox{ km}$.
Then a valid range of distances of observers would be between
$0.00224$ and $50912$ lightyears of $9.4607\cdot 10^{15}\mbox{ m}$ from Earth.
In astronomcal units of $149597870700\mbox{ m}$ the closest distance of an observer would be $141.49 \mbox{ au}$ from Earth.
Due to Earth's oblateness, however, you would get a point spread function significantly different from a dot for this "short" distance from Earth. It might be possible to correct this by an appropriate telescope optics.
The effect of gravitational lensing is
$\theta=\frac{4GM}{rc^2}=2.969\cdot 10^{-27}\frac{\mbox{m}}{\mbox{kg}}\frac{M}{r}$, after applying the constant of gravitation $G$ and the speed of light $c$. With the mass $M=5.97237\cdot 10^{24}\mbox{ kg}$ and a radius of $r=6350000\mbox{ m}$ of Earth, we get an angle of
$\theta=2.793\cdot 10^{-9}$ by gravitational lensing at the surface of Earth.
This would focus parallel rays of light to a point near a distance of
$b=\frac{r}{\tan \theta}=130.27\cdot 10^{15}\mbox{ m}$, or $13.77$ lightyears, hence well beyond the minimum distance where an Arago spot could form. But, of course, the innermost peak of the point spread function would be closer to a circular disc at this larger distance with relevant gravitational lensing.
