m = 10.0 g, the mass of the iron sample
ΔT = 75 - 25.2 = 49.5°C, the decrease in temperature
c = 0.449 J/(g-°C), the specific heat of iron
The heat released is
Q = m*c*ΔT
= (10.0 g)*(0.449 J/(g-°C))*(49.5 C)
= 222.255 J
Answer: 222.3 J (nearest tenth)