If a single layer solenoid of inner radius r1 has n turns along its length, a typical thick solenoid of outer radius r2 is made of m layers of n turns each or m*n total turns. Each carries the same current, so the field is found by integrating the B field equation over the winding volume, using a constant current density.
A Bitter magnet uses n annular copper disks separated by insulating disks, instead of wires. The inner and outer radii r1 and r2 of the annulae are the same as before, that is, the number of turns along the solenoid length is the same but there's only one very thick winding instead of m layers of windings. Axial holes in the disks pass cooling water, cleverly solving one of the difficulties faced in building very high current solenoids. This is still an air core magnet (unless they've added iron) so the usual equations will work if you know the current density in the disks.
I've seen treatments of Bitter magnets in books on high field techniques, but I can't remember any titles right now. Sorry! We can probably figure it out, though. I'd expect the potential drop to be constant across the radius of each disk, in which case the current density is inversely proportional to the length of the path carrying the current. Since path length is [tex]2\pi r[/tex], I'd expect the current density to be inversely proportional to radius. You should be able to integrate the usual solenoid equations with that current density to get the field.