Cache size is $32 \hspace{0.2cm} KB$ and cache block size is $32 \hspace{0.2cm} B$. So,
$\text{Number of sets} = \dfrac{\text{cache size}}{\text{no. of blocks in a set } \times \text{ block size}}$
$ = \dfrac{32 \hspace{0.2cm}KB}{2 \times 32 \hspace{0.2cm}B} = 512$
So, number of index bits needed $= 9$ ( since $2^9 = 512$). Number of offset bits $= 5$ (since $2^5 = 32 \hspace{0.2cm} B$ is the block size and assuming byte addressing). So, number of tag bits $= 32 - 9 - 5 = 18$ (as memory address is of $32 \hspace{0.2cm} bits$).
So, $\text{ time for comparing the data}$ $ \text{= Time to compare the data + Time to select the block in set} \\= 0.6 + 18/10 \text{ ns} \\= 2.4 \text{ ns}.$
(Two comparisons of tag bits need to be done for each block in a set, but they can be carried out in parallel and the succeeding one multiplexed as the output).
Reference: https://courses.cs.washington.edu/courses/cse378/09au/lectures/cse378au09-19.pdf
Correct Answer: $A$