Answer
$\mu_s=0.498$
Work Step by Step
Since the ladder is on the verge of sliding, the net force exerted on it is zero, and so is the net torque.
$\sum F_x=0$, $\sum F_y=0$ and $\sum \tau=0$
The author gives us an excellent sketch of the problem, he also analyzed the reaction force on the left end of the ladder twi its two components.
Thus,
$$\sum F_x=F_{Cx}-F_W=0$$
Hence,
$$F_{Cx}=F_W\tag 1$$
whereas $F_{Cx}$ is the static friction force that is given by
$$F_{Cx}=\mu_sF_{Cy}$$
whereas $F_{Cy}$ is the normal force exerted on the left end of the ladder.
Plugging into (1) and solving for $\mu_s$;
$$\mu_s=\dfrac{F_W}{F_{Cy}}\tag 2$$
We need to find the two reaction forces of $F_W$ and $F_{Cy}$.
Finding $F_{Cy}$;
$$\sum F_y=F_{Cy}-m_{ladder}g-m_{painter}g=0$$
Thus,
$$F_{Cy}=\left(m_{ladder}+m_{painter}\right)g$$
Plugging the given;
$$F_{Cy}=\left( 12+55\right)\cdot 9.8$$
$$F_{Cy}=656.6\;\rm N\tag 3$$
Finding $F_{W}$;
$$\sum \tau_{\text{At left end}}=(F_{C}\cdot 0)-(m_{ladder}g\cdot 1.5)-(m_{painter}g\cdot 2.1)+(F_W\cdot 4 )=0$$
Thus,
$$4F_W=(m_{ladder}g\cdot 1.5)+(m_{painter}g\cdot 2.1)$$
Solving for $F_W$;
$$ F_W=\dfrac{( 12\cdot 9.8\cdot 1.5)+(55\cdot 9.8\cdot 2.1)}{4}$$
$$F_W=327.1\;\rm N\tag 4$$
Plugging (3) and (4) into (2);
$$\mu_s=\dfrac{327.1 }{656.6 } $$
$$\boxed{\mu_s=0.498}$$
