From the question, we see that the mass of the body is 6 kg and the displacement is given by
$x=\frac{t^2}{4}$
So, the velocity would be equal to $v=\frac{dx}{dt}=\frac{t}{2}.$ Thus$ ,v(0) =0 \,and\, v(2) =\frac{2}{2}=1$
Hence the initial kinetic energy would be equal to
$KE{i}=\frac{1}{2}m[v(0)]^2=0$ and similarly the final kinetic energy would be equal to
$KE_f=\frac{1}{2}[mv(2)]^2=\frac{1}{2}\times 6 \times 1^2=3 $
So, we see that the work done by the force is equal to $W=KE_f-KE_i=3-0=3 J$