Since physical address space = 2^16
page size = frame size = 512 bytes
So number of frames = 2^16/2^9= 2^7
Number of bits used to address frame = log2(2^7)= 7 bits
since page table entry is 2 bytes = 16 bits
so now page table entry bit = no of bit used for farme number + 1 valid + 3 bit for protection + 1 dirty bit + Required Aging information
so 16 = 7+1+3+1+x
x=4
is it right ?
They have given answer as 3