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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.

Partial:

From the image i conclude that there are 1 single digit number, 4 two digit numbers, 5 three digit numbers and 1 four digit number. Since they are in order it means-
Single digit- A
Two digit- B,C,D,E
Three digit- F,G,H,I,J
Four digit- K
Now, since H is a three digit number that is both square and palindrome, two possible candidates are 121,484. But it can't be 121 as there need to be two more three digit square numbers(F,G) before H.
Hence, $H = 484$

This is what i have so far.

Partial:

OP clarified there are 4 two digit numbers, 6 three digit numbers and 1 four digit number. Since they are in order it means-
Two digit- A,B,C,D
Three digit- E,F,G,H,I,J
Four digit- K
Now, since H is a three digit number that is both square and palindrome, three possible candidates are 121,484,676. But it can't be 121 as there need to be three 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.

Source Link

Partial:

From the image i conclude that there are 1 single digit number, 4 two digit numbers, 5 three digit numbers and 1 four digit number. Since they are in order it means-
Single digit- A
Two digit- B,C,D,E
Three digit- F,G,H,I,J
Four digit- K
Now, since H is a three digit number that is both square and palindrome, two possible candidates are 121,484. But it can't be 121 as there need to be two more three digit square numbers(F,G) before H.
Hence, $H = 484$

This is what i have so far.