A certain processor uses a fully associative cache of size 16 kB. The cache block size is 16 bytes. Assume that the main memory is byte addressable and uses a 32-bit address. How many bits are required for the $Tag$ and the $Index$ fields respectively in the addresses generated by the processor?

To determine the bits for each field:

1. Block Offset (Byte Offset): The block size is $16\text{ bytes} = 2^4\text{ bytes}$, so the block offset requires:

   $$\log_2(16) = 4\text{ bits}$$

2. Index Bits: In a fully associative cache, any memory block can be placed in any cache line. Effectively, there is only $1$ set containing all lines. Therefore, no index bits are required:

   $$\text{Index bits} = 0$$

3. Tag Bits: Given a $32$-bit physical address:

   $$\text{Tag bits} = \text{Address bits} - \text{Block offset bits} - \text{Index bits} = 32 - 4 - 0 = 28\text{ bits}$$

Hence, the $Tag$ and $Index$ fields require 28 bits and 0 bits, respectively, matching option (D).