The answer is #lambda = 656 nm#.
SInce the electron makes a transition from #n = 3# to #n=2#, we are dealing with a photon emission, which means that the energy difference (#DeltaE#) is actually negative.
#DeltaE = -R_H * (1/n_f^2 - 1/n_i^2)#, where
#n_f# - the final energy level;
#n_i# - the initial energy level;
#R_H# = #2.18 * 10^(-18)J# - Rydberg's constant expressed in J, since energy is expressed in J (this is calculated by multiplying the known #R = 1.097 * 10^7m^(-1)# by #h*c#, Planck's constant times the speed of light).
So, the energy difference for a #n=3# to #n=2# transition is
#DeltaE = -2.18 * 10^(-18) * (1/2^2 - 1/3^2) = -2.18 * 10^(-18) * 0.139 = -3.03 * 10^(-19)J#
We know that the #DeltaE# is equal to the energy of the emitted photon, which is always positive, so
#DeltaE = -h* nu = -(hc)/(lambda) -> lambda = (h*c)/(-DeltaE)#
#lambda = (6.626 * 10^(-34) * 3.00 * 10^8)/(3.03 * 10^(-19)) = 6.56 * 10^(-7)m#
Since we're in the visible range of the EM spectrum (#400-700nm#), the wavelenght could be given as
#lambda = 6.56 * 10^(-7)m = 656 * 10^(-9) m= 656 nm#.
