Solution

Initially, accumulator register is cleared by instruction $CLA$.

So,

$AC=0000$

Next instruction is:

$ADD\ 016$

At location $016$, data is:

$C1A5$

So,

$AC=0000+C1A5=C1A5$

Next instruction is:

$BUN\ 014$

So, control goes to location $014$.

At location $014$, instruction is:

$AND\ 017$

At location $017$, data is:

$93C6$

Now,

$AC=C1A5\ AND\ 93C6$

Convert into binary:

$C1A5=1100\ 0001\ 1010\ 0101$

$93C6=1001\ 0011\ 1100\ 0110$

Now perform bitwise AND:

$1100\ 0001\ 1010\ 0101$

$AND$

$1001\ 0011\ 1100\ 0110$

$=1000\ 0001\ 1000\ 0100$

Now convert back to hexadecimal:

$1000\ 0001\ 1000\ 0100=8184$

So, final content of accumulator is:

$AC=8184$