The wave function for an electron that is confined to x >= 0 nm See the image below. a) What must be the value of b? I don't know if this is correct but is this simply 1/sqrt(L/2) =.559 mm^-1/2 b) What is the probability of finding the electron in a 0.010 nm-wide region centered at x = 1.0 nm? Is this simply .00228630*100 = .23% c) What is the probability of finding the electron in the interval 1.15 nm

Hello!
Psi is the standard symbol for a wave function. Its square is the probability density pd(x). By the definition of probability density, the probability of being between c and d is  int_c^d pd(x) dx. In our case for positive c and d it is
b^2 int_c^d e^(-(2x)/L) dx = b^2*L/2*(e^(-(2c)/L) - e^(-(2d)/L)).
a) the value of b must be such that the total probability, int_(-oo)^(+oo) pd(x) dx, = 1. In our case it is  int_0^(+oo) b^2 e^(-(2x)/L) dx = b^2*L/2 = 1.
So yes, b=sqrt(2/L) and for L=6.4 it is about  0.559 ((nm)^(-1/2)).
And the formula for a probability becomes 
for positive c and d. If c is negative, c must be replaced with zero.
b) use this formula for c=1-0.005 and d=1+0.005.
c) use this formula for c=1.15 and d=1.84.
(there is an error at the picture, must be "for x>=0 nm", not "for x>=nm")