The wave function for a particle is psi(x)=Axe^(-x^2/a^2) where A and a are constants. Where is the particle most likely to be found? Assume that a = 2.49 nm. Is the particle most likely around 2.49 nm because that is the constant/max.

Hello!
The probability of finding a particle within some set R is int_R |Psi(x)|^2 dx. The probability to find a particle at a specific point is zero, but there is a correct question: "what is the point x_0 such that the probability of finding the particle within a small interval with the center in x_0 is maximal?"
Since our Psi(x) is continuous, the integral over a small interval is almost equal to Delta x*|Psi(x_0)|^2. So, we have to find the point(s) x_0 where |Psi(x_0)|^2 has its maximum. This is the same as the |Psi(x_0)| maximum.
The factor |A| has no effect on x_0, thus it is sufficient to find the maximum of f(x) = x e^(-x^2/a^2) for xgt=0 (for xlt0 the values are the same). At x=0 the value is zero, at +oo the limit is also zero, so the maximum is somewhere in between. The necessary condition is f'(x_0) = 0, so the equation is:
f'(x) = (x e^(-x^2/a^2))' =e^(-x^2/a^2) - x*(2x)/a^2e^(-x^2/a^2) =e^(-x^2/a^2)(1-(2x^2)/a^2) = 0.
The only such x_0 = |a|/sqrt(2) (so there are two points of a maximum, |a|/sqrt(2)  and -|a|/sqrt(2)). Numerically for a=2.49 nm   x_0 approx +-1.76 nm.