Answer:
The answer to your question is  C = 0.037 cal/g°C
Explanation:
Data
mass of the wire = m = 237 g
temperature 1 = T1 = 25°C
temperature 2 = T2 = 107°C
Heat = Q = 722 cal
Specific heat = C
Process
1.- Write the formula to find the specific heat
      Q = mC(T2 - T1)
-Solve for C
      C = Q / m(T2 - T1)
2.- Substitution
       C = 722 / 237(107 - 25)
3.- Simplification
       C = 722 / 237(82)
       C = 722 / 19434
4.- Result
        C = 0.037 cal/g°C