What is asked by this question re cache memory?
The question does indeed make some implications which may not be too obvious.

I would infer:


16KB = 4K words ==> 1 word = 4 bytes = 32 bit.
1 block = 4 words ==> 1 block = 4 * 4 bytes = 128 bit

Verify: 16KB / ((4*4 bytes)/block) = 1024 blocks -- check.



For each block(/line) of the cache there is some "administrative" data needed for the logic to know "what's cached where". This is the "tag".
In the given example it is already calculated that for each block (128 bits) of data there will be (32-10-2-2)+1 = 19 bits needed for this administrative information, which means there is an overhead of (19/128) ~ 15% of extra storage required.