The magic words you're probably reaching for are "radiative cooling".
If your ball is in a vacuum far from any star and doesn't generate any heat of its own, this isn't too difficult a problem to solve. There's even a handy calculator for it (and no doubt many others elsewhere) but I'll summarise the key bits here.
The cooling rate in terms of power per unit area is defined by the Stefan-Boltzmann law, and given a nice spherical black body radiator you can work out the radiated power as $$P = 4\pi r^2 \sigma(T_{ball}^4 - T_{ambient}^4)$$
where $\sigma$ is the Stefan-Boltzmann constant. Given that this experiment is being run in deep space and is stopping at merely the freezing point of water, $T_{ambient} \ll T_{ball}$, so you can quietly disregard the ambient term and not go too far wrong.
The hyperphysics calculator suggests using a nice simple model for the energy content of your hot ball as $E = N \frac{3}{2}k_BT$ where $N$ is the number of particles, $k_B$ is Boltzmann's constant and $T$ is the temperature of the object.
Throwing these together and doing a little calculus that I won't try to repeat here gets you a nice simple cooling rate equation:
$$t_{cooling} = \frac{Nk_B}{8\sigma \pi r^2}\left[ \frac{1}{T_{final}^3} - \frac{1}{T_{initial}^3} \right]$$
You can get $N$ from $\frac{mN_A}{M}$ where $m$ is the mass of your sphere, $N_A$ is Avogadro's constant and $M$ is the molar mass of whatever your ball is made of.
For a worked example, lets imagine your sphere is made of some ideal material with the approximately the properties of ruby... density ~4000kg/m3, molar mass ~100g/mol and a magical emissivity of 1. If it were a sphere 1m in radius, it would have mass ~16755 kg and so $N$ is basically 1029.
Throwing all that into the cooling equation gets you a cooling time of about 8 hours.
Obviously using real materials is more problematic. Handling emissivity is easy, but the simple thermal energy model breaks down, especially with awkward stuff like water. As a starter for 10 though, it should do.