Which ONE of the following is the CORRECT option for the sizes of the main memory and the cache memory in this system (byte addressable), respectively?

To determine the sizes of the main memory and the cache memory:

1. Main Memory Address Breakdown:

   • Tag bits = $4$

   • Index bits = $12$

   • Block offset bits: Since the cache block size is $1\text{ byte}$ and the system is byte-addressable, the offset bits $= \log_2(1) = 0\text{ bits}$.

   Total physical address bits:

   $$\text{Address bits} = \text{Tag} + \text{Index} + \text{Offset} = 4 + 12 + 0 = 16\text{ bits}$$

2. Main Memory Size:

   $$\text{Main Memory Size} = 2^{16}\text{ bytes} = 64\text{ KB}$$

3. Cache Memory Size:

   $$\text{Number of cache lines} = 2^{\text{Index bits}} = 2^{12} = 4096 = 4\text{ K}$$

   $$\text{Cache Data Size} = (\text{Number of lines}) \times (\text{Block size}) = 4096 \times 1\text{ byte} = 4\text{ KB}$$

Therefore, the sizes of the main memory and the cache memory are 64 KB and 4 KB, respectively, which corresponds to option (A).