Given,
LA= 36 bits, MM= 30 bits, Page size= 4kB.
Now, number of bytes in a page= Page size/ Byte size = 4kB/1B = 4k(Since,byte addressable).
Thus number of bits required to represent an offset = log(4k) = log(2^2 × 2^10) = 12 bits.
Hence, number of bits for Frame = 30-12=18 bits ~ 3B( 1B=8 bits)
Number of bits for Page = 36-12 =14 bits.
Thus,number of entries in the page table = 2^24 = 2^20 × 2^4 = 16M entries.
Therefore, size of the page table = 16M × 3B = 48MB.