Given, memory address is of 16-bits.
Each cache block contains 16 bytes(i.e. = $2^4$ bytes), so the WORD offset field will be of 4 bits (since memory is byte addressable).
Then we have total 16 blocks (i.e = $2^4$ blocks) in the cache. So BLOCK field will be of 4 bytes.
Then the remaining 16 - 4 - 4 = 8 bits are for TAG field.
TAG(8 bits) | BLOCK(4 bits) | OFFSET(4 bits)
Memory address given: $(9A81)_{16}$
$1001-1010-1000-0001$
So, it maps to block $(8)_{16}$