Method I:

$11 \times 11 = 121$ $111 \times 111 = 12321$ $1111 \times 1111 = 1234321$

Following the same pattern, we get: (1111 ... 9 times) $\times$ (1111 ... 9 times) = 12345678987654321

Sum of the digits of the resulting number = $2 \times (1 + 2 + 3 + .... + 8) + 9 = 2 \times [\frac{8 \times 9}{2}] + 9 = 72 + 9 = 81$

[ Sum of first n natural numbers = $\frac{n (n + 1)}{2}$]

Hence, option (C) is correct.

Method II:

Since each digit of a 9-digit number is 1, so the sum of its digits = 9. So, this number is divisible by 9.

Any multiple of such a number will also be divisible by 9.

So, the sum of the digits of the resulting number of the multiplication (111111111) $\times$ (111111111) must also be divisible by 9.

Therefore, the correct answer will be the option which is a multiple of 9, i.e. 81.

Hence, option (C) is correct.