Allow me to use "[" as absolute syntax.

sqrt((x-3)^2) = [x-3]. The absolute value can not be negative.
Then [x-3] = x-3 if x-3>=0
or [x-3] = 3-x if x-3<=0 (x-3<=0, thus 3-x>=0)
This case the given information cites that [x-3] = 3-x hence x-3<=0 or x<=3

Let's go to the choices.
1. x is not 3, this is insuff to determine whether x<=3
2. -x*[x]>0 so -x>0 or x<0.
If x<0 then surely x will <= 3. Then suff.
So pick B.

I am trying to explain the question first and then will go to the solution

sqrt ( x-3) ^2 = ( 3 - x ) will be true if x - 3 is negative because negative ^ 2 will be positive followed by a sqrt will return positive.

other way round if x- 3 is positive then sqrt of the square of x - 3 will be positive but 3 - x will be negative. hence to Answer whether sqrt ( x-3) ^2 = ( 3 - x ) is True or not. We need to simply find out the sign of x - 3 or in other words if x > 3 or x < 3

Now Statement 1: x not equal to 3 doesn't tell us any thing whether x > 3 or x < 3 ( Note if statement 1 was x = 3 .. this would have been sufficient)

Statement 2 - x | x| > 0 means x has to be negative and hence this tells us that x < 3. or x - 3 < 0.

Hence statement 2 is sufficient to answer the question

IMO B

Let me see if I can simplify this question a little bit more.

Since we know the equation to the left MUST end in a positive number or 0, then the question is really asking "Is (3 - X) ≥ 0", and we can deduce it down even further and say is 3 ≥ X?

(1) says that X is not 3, but it doesn't tell you anything else. Insufficient.

(2) X has to be negative, and therefore 3 must be greater than X. Sufficient.