Find a square number which has two thousand and eighteen $6$s and some numbers of $0$s?
It is impossible.
If the unit digit is $6$, then the tens digit must be an odd number. This is impossible because all digits can be only $6$ or $0$.
If the unit digit is not $6$, then this number must end with even numbers of $0$. Removing these trailing $0$s, we end up in the previous situation (i.e. ending with $6$) which is impossible.