I'll do one trial, you can do the others, and take an average.
Heat lost by hot water = q = mC∆T = (50 g)(4.184 J/g/deg)(26.4 deg) = 5522.9 J
Heat gained by cold water = q = mC∆T = (50 g)(4.184 J/g/deg)(26 deg) = 5439.2 J
The difference would be the heat absorbed by the calorimeter. This is 5522.9 - 5439.2 = 83.7 J
The calorimeter constant would thus be 83.7 J/26 deg = 3.22 J/deg
NOTE: the ∆T for hot and cold is almost the same (26.4 v. 26), so little heat was lost to the calorimeter. This is unusual. You estimated value of 30-85 would be the joules, but the cal constant is J/degree, so the answer here would be 3.22 J/deg.