$a = 11111111.....81\ digits$
Divisibility by 9
If sum of digits of a number is divisible by $9$ then that number is divisible by $9$
So sum of digits of $a =81$ which is divisible by $9$
Hence $a$ is divisible by $9$
Divisibility by 27
Group the digits of $a$ in block of $3$ from right to left i.e. $111 , 111, 111, ....27$ blocks
Now if the sum of these blocks are divisible by $27$ then the number is divisible by $27$
i.e. $111+111+111....27$ terms = $111*27$ which s divisible by $27$
Hence $a$ is also divisible by $27$
Divisibility by 81
Generalized Divisibilty Rule says,
If the number is too large, you can also break it down into several strings with $e$ digits each, satisfying either $10^e = 1$ or $10^e = -1 (mod\ D)$. The sum (or alternate sum) of the numbers have the same divisibility as the original one. (where $D$ is the divisor)
$D=81$
$10^9 \ mod\ 81 = 1 \implies e=9$
So we make blocks of $9$ digits and add them
$111111111+ 111111111+ 111111111+111111111+111111111+111111111+111111111+111111111+111111111 = 999999999$
Now if $999999999$ is divisible by $81$ then $a$ is also divisible by $81$
For checking divisibility of $81$ one method is
Subtract 8 times the last digit from the rest, if the result obtained is divisible by $81$ then the number is also divisible by $81$
We apply this rule recursively to get the answer as follows.
$99999999 - 9*8 = 99,999,927$
$9999992 - 7*8 = 9,999,936$
$999993 - 6*8 = 999945$
$99994 - 5*8 = 99954$
$9995 - 4*8 = 9963$
$996 - 3*8 = 972$
$97 - 2*8 = 81 $
Since $81$ is divisible by $81 \implies 999999999$ is divisible by $81 \implies$ $a$ is also divisible by $81$
$\therefore$ Option $C.$ is correct.
https://en.wikipedia.org/wiki/Divisibility_rule#Beyond_20http://www.vbforums.com/showthread.php?257144-Divisibility