Answer
When the ball hits the floor, the impulse was $0.13~kg\cdot m/s$
Work Step by Step
We can find the speed $v_1$ just before the ball hits the floor when y = 1.5 m.
$v_1^2 = 2gy$
$v_1 = \sqrt{2gy} = \sqrt{(2)(9.80~m/s^2)(1.5~m)}$
$v_1 = 5.42~m/s$
We can find the speed $v_2$ just after the ball hits the floor when y = 0.85 m.
$v_2^2 = 2gy$
$v_2 = \sqrt{2gy} = \sqrt{(2)(9.80~m/s^2)(0.85~m)}$
$v_2 = 4.08~m/s$
If we let the up direction be positive, then $v_1 = -5.42 ~m/s$. We can use the change in momentum to find the impulse.
$Impulse = \Delta p$
$Impulse = m \Delta v = m(v_2-v_1)$
$Impulse = (0.014~kg)(4.08~m/s-(-5.42~m/s))$
$Impulse = (0.014~kg)(9.50~m/s)$
$Impulse = 0.13~kg\cdot m/s$
When the ball hit the floor, the impulse was $0.13~kg\cdot m/s$