We certainly wouldn’t be able to hear it ring - the lack of air in space would keep the sound waves from making it to us, but a cell phone does produce radio waves, and radio waves don’t need air to make it from the moon to the earth.
Let’s say a cell phone produces a radio signal of 1 Watt at a frequency of 1.8 GHz. Your average light bulb produces 60 Watts of energy at a much higher frequency (hence why we can see the energy produced as light!) 1.8 GHz is a radio wavelength, so we’re not going to notice this radio wave coming from our phone. 1 watt is not very much energy, and the moon is pretty far away - but we can work out what the energy would be when it arrives at earth.
We know that as light (or any other wave) radiates away from a point source, the further from the source you are, the weaker the light appears. You can quantify how much weaker it is if you know the distance - power decreases with the distance squared - if you’re twice as far away, the light is four times weaker.
Now, since we know the distance to the moon, the strength of the signal when it starts, we can figure out how much weaker it is when it arrives at the earth.
Radio astronomers use a super intuitive unit of radio power: the Jansky. A Jansky is defined as 10^-26 Watts per square metre per Hertz, which is a super tiny amount of power. So to figure out how many Janskys of power would be arriving at the earth from this single cell phone on the moon, all we have to do is to divide the power produced by the phone by the square of distance to the moon in metres, and then divide by the frequency of the power we’re looking at.
In our case, the power is 1 Watt, the distance to the moon is 384,400 km (3.844 x 10^8 metres), and our frequency is 1.8 GHz, or 1.8 x 10^9 Hz. This gives an equation of:
=1/((3.844 x 10^8)^2) / (1.8x10^9)
Now, nearly 1.5 Janskys of power arriving at the earth is a pretty small amount of power. Can you even detect this little amount of power?
It turns out that you can! Radio telescopes regularly use calibration sources that are about the same brightness as our moon cell phone! And these are relatively bright sources - you want to calibrate to something that is bright enough to see relatively easily. A science target, which is usually much fainter than a calibrator, can be a hundred times fainter than this! Imagine trying to look at a very faint object in the sky, and having a cell phone signal come in from the moon and overwhelm your science target by being 100 times brighter than your target.
So a cell phone on the moon is definitely loud enough to be heard by current radio telescopes, even if there’s no way the sound of the ring would make it to earth.
Have your own question? Feel free to ask!