A 6.5 m long ladder rests against a vertical wall, reaching a...

Let the length of the ladder be $L = 6.5 \text{ m}$ and the height of the wall it reaches be $h = 6.0 \text{ m}$.

The distance of the base of the ladder from the wall is given by the Pythagorean theorem:
$x = \sqrt{L^2 - h^2} = \sqrt{(6.5)^2 - (6.0)^2} = \sqrt{42.25 - 36.00} = \sqrt{6.25} = 2.5 \text{ m}$

The man stands halfway up the ladder. The horizontal distance of the man from the vertical wall (which passes through the upper end of the ladder) is:
$r_{\perp} = \dfrac{x}{2} = \dfrac{2.5}{2} = 1.25 \text{ m}$

The force exerted by the man on the ladder is equal to his weight, acting vertically downwards:
$F = mg = 60 \times 9.8 = 588 \text{ N}$

The torque of this force about the upper end of the ladder is the product of the force and the perpendicular distance from the axis of rotation to the line of action of the force:
$\tau = F \times r_{\perp} = 588 \times 1.25 = 735 \text{ N-m}$

Rounding to two significant figures, the torque is approximately $740 \text{ N-m}$.