ABCD is square of 27 cm.E is point on the side CD such that DE= 20.25

Please refer to the diagrams in the attachment.

Refer to diagram 1: DE measures 20.25. Therefore, E divides DC as 20.25 (DE) and 6.75 (EC) i.e., in the ratio 3: 1.

Refer to digram 2: Extend BC and AE to meet at F.
ABF is a right triangle and the circle drawn touching the 3 sides AB, BC, and AE is nothing but the circle inscribed in this right triangle.
For a right triangle the quickest formula to find the radius of the inscribed circle, r = s - h, where s is the semi perimeter and h is the hypotenuse (I will provide the derivation for this as a reply to this post)
We know one of the sides AB. We have to compute the remaining two sides of the right triangle i.e., hypotenuse AF and side FB.

Triangles ABF and ECF are similar triangles (AA or AAA rule).
Therefore, the ratio of corresponding sides will be same for all 3 corresponding sides.
Of the values available, AB and EC are corresponding sides. Ratio of AB : EC = 27 : 6.75 = 4 : 1
i.e., if the sides of ABF measure 4k units, that of ECF will measure k units.
Therefore, ratio of BF : CF will also be 4 : 1
If CF = x, then (27 + x) : x = 4: 1
Therefore, x = 9.

Hence, BF = 4 * 9 = 36.
AB = 27 (which is nothing but 3*9), BF = 36 (which is nothing but 4*9). Therefore, the hypotenuse will be 5*9 = 45 units.

Semi perimeter of the triangle = (27 + 36 + 45)/2 = 108/2 = 54 units.
Radius of inscribed circle = s - h = 54 - 45 = 9 units.