We are given with a memory which is byte addressable and the size of the memoryis$8$GBandalso size of the byte which is addressable is $4$ byte or we can say that the memoryis$4$Byteaddressable. Now, consider the same memory device with K bit logical address anda$2$KBpagesize and the page table entries are $4$ BYTE each. Assume the memory is againByteaddressable.Calculate:

a$.$ Number of address bits$($bits$).$ b$.$ Number of entries in the page table$($pages$).($divide your answer by $2^(19)$ andreportyourfinalanswer. c$.$ Size of the page table $($MB$).$