Let

d1 = 120 mm be the diameter before the melting

v1 = 1000 mm^3/s the ice melting rate

d2 = 75 mm the second diameter after t2 second of melting

v = the rate of the balls surface area decreasing and is to be found

A = 4 pi r^2 [Surface area of a sphere]

V = 4/3 pi r^3 [Volume of a sphere]

r = radius of the sphere

t1 = 0 - the start moment

r1 = d1 / 2 = 120 / 2 = 60 mm

A1 = 4 pi r1^2

V1 = 4/3 pi r1^3

t2 - the moment when d2 = 75 mm

r2 = d2 / 2 = 37.5 mm

A2 = 4 pi r2^2

V2 = 4/3 pi r1^3

First lets find t2:

Using v1 = (V1 - V2)/(t2 - t1) we can find t2 that is

t2 = (V1 - V2) / v1

t2 = [4/3 pi (r1^3 - r2^3)] / 10^3

Now v, the rate of the balls surface area decreasing is

v = (A1 - A2) / (t2 - t1) [ the difference in the area surfaces divided by the difference in times]

v = [4 pi (r1^2 - r2^2)] / [[4/3 pi (r1^3 - r2^3)] / 10^3]

v = 3 10^3 (r1^2 - r2^2) / (r1^3 - r2^3)

v = 40.31 mm^2 / s approximately

The rate at which the surface area of the ball is decreasing at the moment when it reaches a diameter of 75mm is v = 40.31 mm^2 / s approximately.