Given, F=(ti^+3t2j^)N and m=1kg
Now, we can write from Newton's second law, F=ma=mdtdv
Then, dv=(ti^+3t2j^)dt
Integrating the above,
∫0vdv=∫0t(ti^+3t2j^)dt⇒v=2t2i^+t3j^
Now, power developed by the force is P=F⋅v=(ti^+3t2j^)⋅(2t2i^+t3j^)
=(2t3+3t5)
At t=2s, power is P=(223+3(2)5)=4+(3×32)=100W