The answer to the second virtual memory question is wrong. Actually there is not enough information to answer the question. For 16 bit virtual addresses and 256

The answer to the second virtual memory question is wrong. Actually there is not enough information to answer the question. For 16 bit virtual addresses and 256 byte pages a single level page table will have 256 entries, but how large is an entry? It will be at least one byte because of the status bits but it also has to contain the frame the page is mapped to and there is no information about the number of frames. Assuming physical addresses are not wider than virtual addresses, frames do not overlap and page table entries are multiple of 8 bits wide, a reasonable assumption is that page table entries are 16 bits wide and therefore the page table occupies two pages or 512 bytes.