Unknown Number

(AA)² = (10A+A)²
= (11A)²
= 121A²
= 121 * 6² = 4356
therefore B= 5(odd), 6 at unit place, D = 4

${(AA)}^{2} = DCBA$

Since, the units place digit remains same, so the value of A can be 0, 1, 5, 6.

${11}^{2} = 121$ which is not four digit number.

${55}^{2} = 3025$ the ten's place digit is even.

${66}^{2} = 4356$ the ten's place digit is odd.

Hence it is the correct answer. Hence D = 4.