Answer
$F_h = 17~N$
Work Step by Step
There is a horizontal force $F$ from the wall on the feet. To find this force $F$, we can consider the torque about a rotation axis at the climber's fingers:
$\sum~\tau = 0$
$a~mg-H~F = 0$
$F = \frac{a~mg}{H}$
$F = \frac{(0.20~m)(70~kg)(9.8~m/s^2)}{2.0~m}$
$F = 68.6~N$
The horizontal forces on the fingers are equal and opposite to this horizontal force. We can find $F_h$:
$4~F_h = F$
$F_h = \frac{F}{4}$
$F_h = \frac{68.6~N}{4}$
$F_h = 17~N$
