How to extract lines by words in specific position, not column?

You can use

grep -E '^.{21}A' file

if you want to include cases like A1023, and

grep -E '^.{21}A\>' file

if you want only lines where A appears as an isolated character

NOTE: In the second example the notation \> will match any trailing empty strings.

excerpt from grep man page

The Backslash Character and Special Expressions

The symbols \< and \> respectively match the empty string at the beginning and end of a word. The symbol \b matches the empty string at the edge of a word, and \B matches the empty string provided it's not at the edge of a word. The symbol \w is a synonym for [_[:alnum:]] and \W is a synonym for [^_[:alnum:]].

bash:

while IFS= read -r line; do 
    [[ ${line:21:2} == "A " ]] && echo "$line"
done < file

> awk -v FS= '{ print $22 }' file
A
A
C
A

> awk -v FS= '$22=="A" { print; }' file
ATOM     57  O   LYS A   7       2.254  25.484  18.942  1.00 14.46
ATOM     77  NH1AARG A   8       5.557  19.204  13.388  0.55 24.50
HETATM 1668  O   HOH A1023      25.873  38.343   2.138  1.00 21.99