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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 $(8)_{16}$

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