A computer with 32-bit address uses a two-level page table field and offset. How large are the pages? How much maximum space required when page tables loaded into memory of each entry required 4 byte?

0. 
   If first 8 bit of the address serve as a index into the first level page table and 12 bits are used as offset then 12 bits are used as a index for second page table.
   Now, first page table contains 28 entries and second page table contains 212 entries. Since one page table entries require 4 byte, total space requires to store both table into memory equals to
   =28 ×212 ×4 𝑏𝑦𝑡𝑒𝑠=4194304 bytes