Answer:
147 Kibibits or 18.4 KiB
Explanation:
Given that KiB = kibibyte = 2^10 bytes
Hence 16 KiB is 16384 (2^14) bytes = 4096 (2^12) words.
Therefore, with a block size of 4 words (2^2), we have 1024 (2^10) blocks
Given 32 bits address
We have 4 x 32 = 128 bits of data
The total bits is calculated as (number of blocks) * (data (32 * 4) + tag and validation bits)
Where index part = 10 bits, Offset = 2 bits
2^10 * (4 x 32 + (32–10–2 - 2) + 1)
=> 2^10 * 147 = 147 kibibits
=> 18.4 KiB
