Answer
The second person should apply a force of 100 N at a point that is 2.40 m from the other end of the board.
Work Step by Step
Since the sum of the two upward forces must equal the weight of the board, the second person must apply an upward force of 100 N.
Let's consider an axis of rotation about the center of the board. Since the board exerts no torque about this axis, the magnitude of each person's torque must be equal. Let $r_2$ be the distance from the center of the board that the 100 N force is applied.
$(100~N)~r_2 = (60~N)(1.50~m)$
$r2 = \frac{(60~N)(1.50~m)}{100~N}$
$r_2 = 0.90~m$
Note that 0.90 m from the center is a distance of 2.40 m from the other end of the board. The second person should apply a force of 100 N at a point that is 2.40 m from the other end of the board.
