Firstly on the basis that your measurement is US gallons, the volume of the sludge that was de-watered is 7,667 m3.

Unfortunately you do not tell us the solids content of the original sludge before de-watering, or the density of the sludge.

If this is activated sludge, and if the un-dewatered sludge was at say 2% solids, and it is now de-watered it to 29.4% solids the mass of de-watered sludge can be estimate as follows:
7,667 m3 at 2% solids = 153,340 kg sludge (dry).

Therefore at 29.4% solids the mass of de-watered cake would be 521,547 kg sludge cake, (i.e. 521.547 metric tonnes).

There are many "if's" in this estimate, but I hope you can follow the reasoning.