A ball of mass 0.15 kg is dropped from a height 10 m, strikes the ground and rebounds to the same height. The magnitude of impulse imparted to the ball is (g = 10 m/s²) nearly :

Velocity while hitting the ground
$\mathrm{v}=\sqrt{2\times 10\times 10}=10\sqrt{2}$
As it goes to the same height it will return with same speed.
So change in velocity v-(-v) = 2v
Change in momentum or impulse
= 2mv
$=2\times 0.15\times 10\sqrt{2}=3\sqrt{2}=4.2\mathrm{kgm}/\mathrm{s}$