Partial:
From the image i conclude thatOP clarified there are 1 single digit number, 4 two digit numbers, 56 three digit numbers and 1 four digit number. Since they are in order it means-
Single digit- A
Two digit- BA,B,C,D,E
Three digit- FE,F,G,H,I,J
Four digit- K
Now, since H is a three digit number that is both square and palindrome, twothree possible candidates are 121,484,676. But it can't be 121 as there need to be twothree more three digit square numbers (E,F,G) before H.
Hence, $H = 484$ or $676$
Also, B is triangular square number of two digit, only one that fits is 36
So, $B=36$
This is what i have so far.