Verified answer

I have no clue how you worked this. But I do see some issues. A high jumper has a not inconsiderable vx = V cos(theta) X direction velocity when jumping. And your problem gives no info to calculate vx. vx would add to the impact kinetic energy when landing on that mat.

1/2 mvy^2 = mgh; vy^2 = 2gh, where vy is the vertical velocity vector due to falling h = 2.13 - .3 m straight down to the mat's surface which is .3 m above ground level.

Without vx, we must assume vx = 0, which is not very realistic. In any case, the mat absorbs the kinetic energy KE = 1/2 mvy^2 = 1/2 m(2gh); where m = 65 kg, g = 9.81 m/sec^2, and h = 2.13 - .3 = 1.83 m

The mat does the following work to absorb KE, WE = 1/2 kdX^2 = 1/2 mvy^2 = 1/2 m2gh = KE, from the coservation of energy. (Note, I had to assume the mat acts like a spring when compressing. But, again, that is not very realistic. Since the mat's compression characteristics were not given, the spring model is a WAG.)

Solve for k, the compression constant for the mat. k = 2mg/dX^2. Then the average force F/2 = kdX/2 = mgh/dX = 65*9.81*(2.13-.3)/.18 = 6482.775 Newtons.

Given the assumptions, I agree with the textbook answer. However, the problem discounts a lot of horizontal energy/velocity that really should not be discounted in a realistic problem.

F = mgh/c where c is the amount the mat compresses.

If you use 2.13 for h, you get ~7500 N

If you use (2.13-.3), you get ~6500 N

Hmm - he hits the mat at velocity sqrt(2 * 9.8 * (2.13 - 0.3)).

the force has to decelerate this velocity to zero in .18m

= this velocity ^2 /(2 * 0.18)

in other words, 2* 9.8 * (2.13 - 0.3)/ 2 * 0.18

Maybe you forgot the height of the mat!