The correct option is A 83 J
Given, →F=3x2^i+2y^j N
Initial position, −−→(r1)=(2^i+3^j) m⇒(2,3)→(x1,y1)
Final position, −−→(r2)=(4^i+6^j) m⇒(4,6)→(x2,y2)
Since, the given force is variable in x and y, displacement of object, →ds=dx^i+dy^j
Then, work done W=∫→F.→ds=∫x2x1Fxdx+∫y2y1Fydy=∫423x2dx+∫632ydy=x3|42+y2|63=(43−23)+(62−32)=(64−8)+(36−9)=83 J