What will be the size of page table in this case?

Question

I was asked this question in a test and I couldn't answer this:

If size of physical memory is 64MB and the virtual address bit space is 32 bits, and the size of each page is 4 KB, then what will be the size of the page table?

How do I do it? I am sort of confused. First I divided, (2^32)/(2^20). This gives me the total number of virtual pages. After this, what do I do? Do i Just multiply it with 4 KB?

Answer 1

There is no answer to that question with the fact given The size of the page table is limited by the virtual address space. If the kernel space is not paged, it is limited by that as well. However, the table table can be any sized within those limits.