why 16KB = 4K words?

I realize that 16KB = 8K words

because 1 word = 2 bytes

Bits in a Cache (An Example) How many total bits are required for a direct-mapped cache with 16 KB of data and 4-word blocks, assuming a 32-bit address? 16KB-4K words-212 words and with a block size of 4 words (22) 210 blocks. Each block has 4X32-128 bits of data plus a tag, which is 32-10-2-2 bits, plus a valid bit. Thus, the total cache size is 210X(128+(32-10-2-2)+1)-210X147-147Kbits or 18.4KB for a 16KB cache. For this cache, the total number of bits in the cache is about 1.15 times as many as needed just for the storage of the data. Chap. 5