One crude method of determining the size of a molecule is to treat the molecule as an infinite square well with an electron trapped inside, and to measure the wavelengths of emitted photons. If the photon emitted during the n = 2 to n =1 transition has wavelength 1940.0 nm, what is the width of the molecule? Is it simply sqrt((3hlambda)/(8mc)) = 1.3nm

Hello!
The formula seems correct. It can be obtained from the following facts:
E_n = (n^2 h^2)/(8mL^2)  and  lambda = (hc)/(E_2-E_1),
where n is the state number (changes from 2 to 1 in our case),h is the Planck's constant,m is the mass of electron,L is the size of a molecule,lambda is the photon's wavelength.
Therefore E_2-E_1 =((2^2-1^2) h^2)/(8mL^2) = (3 h^2)/(8mL^2), and
L^2 = (3h^2)/(8m(E_2-E_1)) = (3h^2)/(8m(hc)/lambda) = (3h lambda)/(8mc).
Thus L = sqrt((3h lambda)/(8mc)). To find the numerical result, recall the values in standard units:
h = 6.6*10^(-34),  m = 9.1*10^(-31),  c = 3*10^8,  lambda = 1.94*10^(-6).
So the result is
sqrt((3*6.6*1.94)/(8*9.1*3)*(10^(-34)*10^(-6))/(10^(-31)*10^8)) approx sqrt(0.176*10^(-17)) = sqrt(1.76*10^(-18)) approx 1.33*10^(-9).
This value is in standard units, meters. Because nano means 10^(-9), this is the same as 1.33 nm. So your numerical answer is also correct:)