#### Answer

The force of friction that the driveway exerts on the bottom of the ladder is $170~N$

#### Work Step by Step

We can find the angle $\theta$ the ladder makes with the wall:
$cos~\theta = \frac{4.7}{5.0}$
$\theta = cos^{-1}(\frac{4.7}{5.0})$
$\theta = 19.95^{\circ}$
We can consider the torques about the rotation axis at the bottom point of the ladder.
The weight of the ladder exerts a clockwise torque about this axis.
The weight of the painter exerts a clockwise torque about this axis.
The normal force $F_N$ from the wall on the top of the ladder exerts a counterclockwise torque about this axis..
Since the system is in equilibrium, the counterclockwise torque is equal in magnitude to the sum of the clockwise torques. We can find the normal force $F_N$:
$\tau_{ccw} = \tau_{cw}$
$(4.7~m)(F_N) = (2.5~m)(120~N)~sin~19.95^{\circ}+ (3.0~m)(680~N)~sin~19.95^{\circ}$
$F_N = \frac{(2.5~m)(120~N)~sin~19.95^{\circ}+ (3.0~m)(680~N)~sin~19.95^{\circ}}{4.7~m}$
$F_N = 170~N$
The normal force from the wall on the top of the ladder has a magnitude of $170~N$. The only other horizontal force is the force of friction which opposes this normal force. Since the system is in equilibrium, the force of friction must have a magnitude of $170~N$.