Ice formation in a cavity that was sealed at ambient condition and taken to space

The ambient temperature may be near $0K$, but the temperature of the device will not be that low, as it is still going to absorb heat from the sun (assuming it is still in the solar system), in fact, it will actually be very hot, rather than being cold.

The heat list through radiation is lower (at low temperatures) than the heat gained. The heat gain and loss will come to an equilibrium at a higher temperature than the freezing point of water

The heat loss by radiation depends on temperature, and is given by the formula: $$frac{Q}{t}=sigma{e}AT^{4}$$

Where $sigma$ is the Stephan-Boltzman constant. And, $ e $ is the emissivity of the object.

At equilibrium,

$$ Q _{absorbed} = Q _{radiated} $$

So,

$$ {1366.1} times {e} times A = 5.67 times 10^{-8} times e times A times T^4 $$

Solving for T (Area and e get canceled on both sides), we get,

$$ T = 393.98 K $$

Also, I suggest you read the article here.