Calculate Paging System
Ok, let's break it down:


Physical: 232
Page Size: 210
Number of Pages: 216


How many bits in a logical address?  That's the page address bits plus the number of pages bits. The upper portion of an address is the page number (16 bits), and the lower portion is the offset within that address (10 bits), so the whole address size is 26 bits (1026 bytes).
How many bytes are in a frame?  A frame is where a page can be mapped into memory, so a frame has to be the same size as a page - 210 bytes.
How many bits are in the physical address specifying the frame?  Well, you have 32 bits of physical address, and a frame is 210 big, so that leaves 22 of the bits (32 - 10) for the frame's base address.
How many entries in the page table?  The page table is the full list of pages, whether mapped or unmapped - so there are 216 entries in the page table, since there are 216 pages.
How many bits in each page table entry? Assume each page table entry contains a valid/invalid bit. If each page maps to an entry in the page table, and that table is a list of the addresses at which the pages are mapped in physical memory, then each address in the table must be the size of answer 3 (22 bits) plus one bit for the valid/invalid bit, so 23 bits.



Is that clearer?