Given, the mass of the body = m

The height from which the body dropped = h

The speed of the body when reached the ground, $${v_f} = 0.8\sqrt {gh} $$

Initial velocity of the body, v = 0 m/s

Using the work-energy theorem,

Work done by gravity + Work done by air-friction = Final kinetic energy $$-$$ Initial kinetic energy.

$${W_{mg}} + {W_{air - friction}} = {1 \over 2}mv_f^2 - {1 \over 2}mv_i^2$$

Here, work done by gravity = mgh

$$ \Rightarrow mgh + {W_{air - friction}} = {1 \over 2}m{(0.8\sqrt {gh} )^2} - {1 \over 2}m{(0)^2}$$

$$ \Rightarrow {W_{air - friction}} = {{0.64mgh} \over 2} - mgh$$

$$ \Rightarrow 0.32mgh - mgh = - 0.68mgh$$

The value of the work done by the air friction is $$-$$ 0.68 mgh.