A level help

Hi can someone help with this physics question. I’m struggling to get to an answer

3C48 is a quasar in the constellation Triangulum. It is believed to have a power output 4x10^11 times great than the Sun. At the Earth, the Suns intensity is 1.4x10^17 times greater than that of the quasar.
Calculate using the inverse square law the distance from Earth to this quasar in AU

Thanks a bunch

Hi Zoeva123
You're right! This one is pretty hard, are you sure its on an a level physics paper? Anyway I tried it for a bit and have managed to get a ratio of their distances (I may be wrong) which is AU (right?).
Anyway:
(Intensity = Work done/Area) and (the inverse of intensity^2 is proportional to the distance "r")
As we are given in the question P(power output) of Sun = P
P(power output) of Quasar = (4x10^11)P
And also I(intensity) of Sun = (1.4x10^17)I
I(intensity) of Quasar = I
If we do (Power Output)/(Intensity) to find the respective (m^2) and then root then we perhaps should have a correct ratio?
That would mean Sun(m^2) : Quasar (m^2) = (7.1428... x 10^-18) : (4x10^11)
Root both = Sun(m) : Quasar (m) = (2.6726... x 10^-9) : (6.324...x10^5)

Now lets divide both by the distance of the sun and then I suppose you have it in Arbitrary Units in terms of the Sun's distance from earth.

So obviously Sun = 1 Sun Unit
And the Quasar = 2.3664...x10^14 Sun Units

This could be quite wrong but hopefully someone can verify this working out!