#### Answer

b)

#### Work Step by Step

We know that the work done is given by:
$$ W = Fd cos \theta$$
The distance is the same, so we compare $Fcos \theta$. In the first case, the angle is 0 degrees and the force is F, so we get a work of F times the distance. In the second case, we have a force of 2F and an angle of 45 degrees. When plugged into the equation, this simplifies to $\sqrt{2} F$ times the distance, which is greater than the work done in the first scenario.