To answer this question, we need to understand the representation of a COMP-3 field in COBOL.
A S9(7) COMP-3 field is a signed numeric field that can hold up to 7 digits. It is stored in a packed decimal format, where each digit is represented by half a byte (nibble). The left-most nibble is used to store the sign of the number.
In a COMP-3 field, each digit occupies 4 bits (half a byte). Therefore, for a S9(7) COMP-3 field, we have 7 digits, which requires 7*4 = 28 bits.
Since 8 bits make up 1 byte, we need to divide 28 bits by 8 to get the number of bytes required.
28 bits / 8 = 3.5 bytes
Since we cannot have half a byte, we need to round up to the nearest whole number. Therefore, a S9(7) COMP-3 field occupies 4 bytes.
So, the correct answer is A) 4 bytes.