A=πr2, dA/dt=k where k is a constant, because rate of change of A is constant.
A=kt+C where C is a constant of integration, to be found.
4π=2k+C, 9π=3k+C, 5π=k, so k=5π (by subtracting these two equations), and 4π=10π+C, C=-6π (by substituting k=5π). We now know how A and t are related:
A=5πt-6π. But A=πr2 so we can relate r and t:
πr2=5πt-6π, r2=5t-6.
When t=2, r2=4, r=2, and when t=3, r2=9, r=3 therefore r changes from 2 to 3, an increase of 1 as t goes from 2 to 3.
Another way of looking at it (without using calculus) is to look at the growth rate of A in the time interval of 3-2=1:
9π/4π=9/4. The growth rate of r is the square root of this because A is proportional to r2, making r proportional to the square root of A. Therefore r changes from 2 to 3 as A changes from 4π to 9π. But this only means an increase of 50%, rather than a net increase (an actual value). But r2=A/π, so when A=4π, r2=4, r=2; when A=9π, r2=9, r=3; which tells us that r actually changes from 2 to 3 when A changes from 4π to 9π, confirming a net increase of 1.