The correct option is (D) 28 kg. m/s.
Given Data:

Mass of the ball, m = 2 kg
When the ball strikes the floor, the velocity of the ball, v₁= 8 m/s
When the ball rebounds upward, the velocity of the ball, v₂= -6 m/s
Momentum of a particle (p) is equal to the mass of the particle (m) times velocity of the particle (v).
So, p = mv …… (1)
When the ball strikes the floor having mass 2 kg (m=2 kg) and velocity 8 m/s (v₁=8 m/s), the momentum of the ball (p₁) will be,
p₁= mv₁
= (2 kg) (8 m/s)
= 16 kg. m/s ……. (2)

When the ball rebounds upward having mass 2 kg (m=₂ kg) and velocity 6 m/s (v₁= -6 m/s), the momentum of the ball (p₂) will be,
p₂= mv₂
= (2 kg) (-6 m/s)
= -12 kg. m/s …… (3)
So the magnitude of the change in momentum will be,
Change in momentum = p₁- p₂
= 16 kg. m/s – (-12 kg. m/s)
= 28 kg. m/s …… (4)
From equation (4) we observed that, the magnitude of the change in momentum of the ball will be, 28 kg. m/s. Thus option (d) is correct