(a) Since a Megaparsec is about 3.1 * 10^22 m, this burst's luminosity is

   L = 4 * pi * r^2 * F = 4 * 3.14 * (3000 * 3.1 * 10^22 m)^2 * (5 * 10^-10 W/m^2)
     = 8.7 * 10^43 W  or 8.7 * 10^43 Joule/sec

   For a 90 second period, the total energy release was

   E = (8.7 * 10^43 Joule/s) * (90 s) = 7.8 * 10^45 J.

(b) The luminosity of a galaxy is usually about equal to the luminosity of all
its stars put together.  Since the Milky Way has about 10^11 stars, if we use
the Sun as a typical star then our galaxy's luminosity is

   L = 10^11 * (4 * 10^26 W) = 4 * 10^37 W.

The luminosity of the gamma-ray burst is 

   (8.7 * 10^43 W)
   --------------- = 2.2 million times greater than an entire galaxy!
    (4 * 10^37 W)

(c) The energy released by the burst was (7.8 * 10^45)/(10^44) = 78 times 
greater than that of a typical supernova.