To answer this question, you need to understand the concept of COMP fields in COBOL.
In COBOL, the COMP field is used to store numeric data. The size of a COMP field is determined by the number of bytes it occupies in memory.
In this case, the S9(8) represents a signed numeric field with a length of 8 digits. The "S" indicates that the field is signed, and the "9(8)" indicates that it can hold 8 digits.
In COBOL, the COMP field is typically represented using the binary format, which means that each digit occupies 4 bits. Therefore, to calculate the number of bytes required for the field, we divide the number of digits by 2.
In this case, the S9(8) COMP field has 8 digits, so it will occupy 8/2 = 4 bytes of memory.
Based on this explanation, the correct answer is B) 4.