a.       How many 128K´16 DRAM (1 word = 16 bits) chips are needed to provide memory capacity of 1M bytes?

b.      How many address lines are required to access 1M bytes? 

a.       The size of one memory is 128´2¹⁰´2⁴ = 2²¹ bits.

The total size of memory needed is 1M bytes = 2²⁰ bytes = 2²⁰´2³ bits = 2²³ bits.

Therefore, we need 2²³/2²¹ = 4 128K´16 DRAM chips to provide memory capacity of 1M bytes.

b.      If each word has 16 bits, then we need log₂(128´2¹⁰) = 17 address lines for each 128K´16 DRAM chip. For the 1M bytes capacity, we need 4 of these chips, i.e. in total 17´4 = 68 address lines. But this is not enough! We still have to select which chip to use i.e. we need 2 extra address lines since log₂(4 chips) = 2. In total we need 70 address lines to access the 1M bytes.

(note that a 2-to-4 decoder is needed to select which chip to use).