Answer
$$0.216$$
Work Step by Step
The forces acting on the climber are
1. The normal force $F_{N1}$ on his feet exerted by the ground
2. The normal force $F_{N2}$ on his hand exerted by the wall
3. The static frictional force $f_s=\mu_sF_{N1}$
4. The weight of the climber $W$
At the static equilibrium, the net force acting on the the climber zero.
Therefore,
at horizontal equilibrium: $F_{N2}-f_s=0\implies f_s=F_{N2} \;.............(1)$
at vertical equilibrium: $F_{N1}-W=0\implies F_{N1}=W\;...............(2)$
The net torque about the contact point between the ground and his feet and is also zero. If $\theta$ be the angle which is made by the climber with the horizontal, then
$Wd\cos\theta-F_{N2}L\sin\theta=0$
or, $F_{N2}=\frac{Wd\cos\theta}{L\sin\theta}$
or, $F_{N2}=\frac{Wd}{L}\cot\theta\;...............(3)$
Now the coefficient of the static friction is given by
$\mu_s=\frac{f_s}{F_{N1}}$
or, $\mu_s=\frac{F_{N2}}{W}$ (using Eq. $1$ and Eq. $2$)
or, $\mu_s=\frac{d}{L}\cot\theta$ ((using Eq. $3$ )
or, $\mu_s=\frac{d}{L}\frac{a}{\sqrt {L^2-a^2}}$
Substituting the given values
$\mu_s=\frac{0.940}{2.10}\frac{0.914}{\sqrt {(2.10)^2-(0.914)^2}}$
$\boxed{\mu_s=0.216}$
Therefore, the coefficient of static friction between feet and ground is $0.216$
