But your logic is still not sound. The sun is 332900 times the mass of 'Terra'. The distance would need to be about 685 AU away to be about the same apparent magnitude as the Sun.
Starting with this refined formula: Gotten from
heremstar = − 2.72 − 2.5 · log(Lstar/diststar^2)
I put -26.73 (the apparent magnatude of the sun as we see it at 1AU) in as mstar
Then put 8700000 in for Lstar (the luminosity of R136a1)
It will give the distance (diststar) for a set star with the given luminosity and apparent magnatude.
-26.73=-2.27 -2.25*log(8700000/diststar^2)
-26.73+2.27=-2.25*log(8700000/diststar^2)
-24.46=-2.25*log(8700000/diststar^2)
-24.46/-2.25=log(8700000/diststar^2)
10.87=log(8700000/diststar^2)
10^10.87= 8700000/diststar^2
74131024130.09=8700000/diststar^2
diststar^2=8700000/74131024130.09
diststar^2= 0.0001173958
diststar= 0.0001173958^-2
diststar= 0.01083327 Light-years.
diststar=63,241.1(AU)*0.01083327 (Ly)
diststar=685 AU
(And sorry I am bad at short cutting my zeroing out of items)
I am getting that nice upper feeling that I get when doing math like that.