Let upward be the positive direction. The initial velocity is $v_i = -5$ m/s, and the final velocity is $v_f = 3$ m/s. During the collision, the net force on the ball is the sum of the average normal force from the ground $\bar{F}_N$ and gravity $-mg$. The impulse-momentum theorem gives $(\bar{F}_N - mg)\Delta t = m(v_f - v_i)$. The average normal force exerted by the ground on the ball is:

$$\bar{F}_N = \frac{m(v_f - v_i)}{\Delta t} + mg$$

By Newton's third law, the force exerted by the ball on the ground has the same magnitude.

$$\bar{F}_N = \frac{0.02 \text{ kg}(3 \text{ m/s} - (-5 \text{ m/s}))}{0.02 \text{ s}} + (0.02 \text{ kg})(9.8 \text{ m/s}^2)$$