No. of 1K x 8 chips required = $\frac{16K * 16}{1K * 8}$
= 16 * 2 chips ( Each row has two 1K*8 chips and a total of 16 rows)
Now, in order to access this bigger memory we need to choose one of the 16 rows.
Let us see how many address lines are required :
Total memory = 16K * 16
= 16K words ($\because$ 1 word = 16 bits here)
= $2^{14}$
Hence 14 address lines are needed to access the memory. Let the address be represented by A13A12...A1A0
We can observe that we need 10 address bits to access each of the 1K * 8 Chip
Remaining Address bits = 14 - 10
= 4 Address bits
Let us suppose we use the least 10 bits to address 1K*8 chip (Since it has 1024 words, we need 10 bits).
We can implement the decoders using the Higher 4 bits as shown in the picture (Picture is self Explanatory)