Solution

The correct answer is 2²⁰.

Key Points

• According to the question, the total size of the main memory is 16 GB. The size of the virtual address space is 32 bits. The page size is 4KB. It is also given that the frame size and page size are the same for the given machine. 
• So, the virtual memory size is = 2³² Bytes.
• frame size = page size= 4KB= (2² x 2¹⁰) Bytes = 2¹² Bytes.
• So, the number of pages required for the given virtual address space is = (2³²/ 2¹²) = 2²⁰. 
• Frame size = Page size = 2²⁰.

Thus the correct answer is 2²⁰.

Additional Information

• The page is referred to as a continuous memory block with a fixed memory size in an operating system. But it is virtual memory. The page is the smallest data set unit in the virtual memory.
• The frame in the operating system is referred to as a fixed length of the physical block. The frame is often found on the RAM block.