To solve this question, the user needs to have an understanding of the different USAGE clauses in COBOL and the storage requirements for each of them.
A. Comp: This usage clause is a binary format that occupies 2 bytes of storage. Therefore, if the data item Var1 is described with PIC S9(4) and its usage clause is Comp, it would require 2 bytes of storage, which is less than the given 3 bytes. Hence, Option A is incorrect.
B. Comp-1: This usage clause is a packed decimal format that occupies 4 bytes of storage. Therefore, if the data item Var1 is described with PIC S9(4) and its usage clause is Comp-1, it would require 4 bytes of storage, which is more than the given 3 bytes. Hence, Option B is incorrect.
C. Display: This usage clause is an alphanumeric format that occupies 1 byte for each character of the data item. Since the data item Var1 is described with PIC S9(4), it would require 4 characters to store its value. Therefore, if the usage clause is Display, it would require 4 bytes of storage, which is more than the given 3 bytes. Hence, Option C is incorrect.
D. Comp-3: This usage clause is a packed decimal format that occupies 3 bytes of storage. Therefore, if the data item Var1 is described with PIC S9(4) and its usage clause is Comp-3, it would require 3 bytes of storage, which matches the given requirement. Hence, Option D is correct.