It depends what you mean by cost. The total cost of the Manhattan Project was $2 billion. So if you apportion half of that to Hiroshima, that's $1 billion, far more than the cost of the ordnance dropped on the single Tokyo raid..
But only 10% of that cost was actually for the construction of the bombs, so again, 10% of $1 billion is $100 million.
There were multiple Tokyo raids, from the small Doolittle raid in 1942,
I assume the Raid you are referring to though is Operation Meetinghouse, March 9–10 1944, during which 1665 tons of bombs were dropped by 282 B-29s that actually made it to the target. I was not able to find the cost of the ordnance used, but if you divide $100 million by 1665 tons, you get about $60 thousand dollars per ton. These were mostly 500 pound bombs, so that would mean $15,000 per bomb in order to equal the cost of the Hiroshima bomb.
I doubt those conventional incendiary bombs cost $15k each, but I could be wrong. Someone more familiar with WW2 ordnance cost may know more, but I seem to recall that torpedoes cost around $20k, and they are far more complicated, obviously.
Again, I’d welcome comments. I think my methodology is solid, just not sure about the unit cost of an E-46 cluster bomb, or the smaller and cheaper 100 pound M47 incendiaries.
Fun Fact: The overall cost of the Manhattan Project, that developed atomic weapons, was $2 billion dollars ($27 billion 2016 dollars)
The plane that carried the bombs, the Boeing B-29 Superfortress, had an overal development cost $3 billion dollars ($41+ billion 2016 dollars)
So the plane development cost more than the bomb development (some estimates put B-29 costs at $3.7 billion ($50+ billion 2016 dollars), and was the most costly single weapon development project in WW2.