Well, if one would like to be technical:
1 J/°K ≈ 10^23 bits of information (
Source)
1 GB ≈ 1.07*10^-14 J/°K = S
E = mc^2
ST = mc^2
Assuming room temperature encoding process at 295°K:
E = S*295°K ≈ mc^2
E/(c^2) ≈ 3.52*10^-29 g
1 ounce = 28.3495231 g
So,
1 GB ≈ 1.24*10^-30 oz
Now we want to find how many GB are in 6 oz:
x ≈ 6 oz / 1.24*10^-30 oz/GB
x ≈ 4.83*10^30 GB
So there're approximately
4.83*10^30 GB in 6 pure ounces of information.
Now, to put that in perspective, with a 100Mbps Fast Ethernet connection, this would take:
4.802012046222223e+27 DAYS
or 1.31561973869e+25 YEARS That's 1 septillion years