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If we look at a number line, we can start at 11 and count back 8 spaces to arrive at 3. However, which choice is equivalent to 3?

Obviously, the choice $\text{Two}$ can be eliminated, as $2\neq 3$

$201^4$ has a units digit of 1, and therefore is equal to $1\pmod 5$ , and since $1\neq 3$ , this choice can also be eliminated.

For the summation choice,

$\sum_{i=1}^{7}i(-1)^{i+1}$

$\implies (-1)^2+2\cdot (-1)^3+3\cdot (-1)^4\ldots$

$\implies 1 - 2 + 3 - 4 + 5 - 6 + 7 = 4$

And since $3\neq 4$ , this choice is clearly not correct.

So, obviously, the last choice, $6-\sqrt{9}$ , must be correct. Testing it, we find that $6-\sqrt{9}=6-3=\boxed{3}\ \blacksquare$ , so that is our desired answer.

6-3=3

there is no difficulty at all...we are ver y sure that the answer is 3 we just have to be confident here and the option 6-sqrt(9) will fetch us the right answer....

9^1/2 =23 So, 6 - 9^1/2 =3. And, 11 - 8 is also = 3.. So the ans. is '6 - 9^1/2'