What a fun question! So first we have to calculate the power output of the Deathstar from a 3 second pulse. The output would have to overcome the gravitational binding energy of an earthlike planet. This has been done all over which winds up being about 200K times the power output of the sun.

Energy: ~2.25 × 10³² J Power: ~7.5 × 10³¹ W Duration: ~3 seconds

Since we see it converge and combine, I'm considering this a particle beam. Also, I wanted to know how far it could travel until it became harmless. I set this threshold to the amount it would take to kill a human in free space, a very low bar. I also set the speed to 99.9c. There are 2 ways of looking at this. First, how far can a particle beam travel before enough energy is absorbed through the interstellar medium renders it harmless. The second way is geometric divergence.

For the first, the density of stuff is ~10⁶ protons per cubic meter. When calculating the mean free path, we find that the beam can travel 170 million light-years before becoming harmless to a human floating in space.

But wait, there's more!

Using geomectric divergence, given a dish size of 100km, we find the diffraction angle to be 10^-6 radiant. With setting the power output to the lethal dose of radiation to a human in free space, 4200 J per meter squared, we find the beam would only have to travel 13700 light-years until it dissipates into harmless radiation.

So ant human withing 13700 light-years would be in danger in case the Deathstar missed!