A pebble is dropped in still water, forming a circular ripple whose
radius is expanding at a constant rate of 10 cm/sec. I need a formula
giving the area enclosed by the ripple as a function of time. When the
radius is 20 cm, and how fast the area enclosed by the ripple
increased?

At time t seconds, the radius of the circle is 10t cm.
At time t seconds, the area is therefore pi x (10t)^2 cm^2 = 100 pi t^2 cm^2

To find the rate of increase in area with respect to time, we need to
differentiate:

d/dt (100 pi t^2) = 200 pi t (cm^2/s)

At time 20 s, the area is increasing at 4000 pi cm^2/s

Oops - the second part was when r = 20.

That will be at time t = 2s.

The rate of expansion will be 400 pi cm^2/s

Do you know what rule you used to find the derivative?

Yes.  It's differentiation of a polynomial in x.

For each term, d/dx ax^n = anx^(n-1).
For example, d/dx 2x^4 = 2*4x^3 = 8x^3.

In this case, it was d/dt 100*pi t^2 = 200*pi t