Final answer:
The cache word size would be 18 bits based on the given information, but since that is not an option, the correct choice is (d) None of these.
Step-by-step explanation:
When discussing the size of a cache memory word for a direct-mapped cache system, it is essential to consider both the data part and the additional bits required for the address tag. In this scenario, for an 8k-byte main memory and a 2k-word cache memory using direct mapping, we first calculate the number of bits required to index into the cache.
Since it's a 2k-word cache, there are 2^11 words, which means we will need 11 bits for the index. The size of the main memory tells us the addressable units, which is 2^13 bytes (8k bytes = 8192 bytes). Hence, we would need 13 bits to address each byte. As the cache is 2k words and each word is likely the same size as the main memory's addressing unit, this might suggest that a word is a byte, but the question does not provide the information necessary to confirm this.
However, we also need to consider the tag size that would be a part of the cache word. The tag size is calculated by subtracting the index bits and the byte offset bits (if we assume a byte-addressable memory and each word is one byte) from the total address bits. Assuming that each word in cache corresponds to a byte in memory, and there are no byte offset bits (since the question does not imply larger word sizes), this leaves 13 (total) - 11 (index) = 2 bits for the tag. Therefore, each word of cache memory comprises of the data itself (16 bits if we assume a word size of 2 bytes), plus the tag size (2 bits), totaling 18 bits. Thus, the correct answer with the information given is (d) None of these.