The energy of the photon emitted shall equal to the difference between the electron's potential energy at the two orbits.
The ul("Rydberg Formula") suggests that an electron that travels (or literally "falls") from energy level n_i to n_f (n_i>n_f) emits a photon of wavelength lambda for which
1/lambda=R(1/n_f^2-1/n_i^2)
where the Rydberg's Constant R=1.097 xx 10^7 color(white)(l) m^{-1}.
In this particular scenario
Such that
1/lambda = R(1/n_f^2-1/n_i^2)
color(white)(Delta "PE") = 1.097 xx 10^7 color(white)(l) m^(-1)* (1/1^2-1/3^2)
color(white)(Delta "PE") = 9.751*10^7 color(white)(l) m^(-1)
lambda = 1.026 xx 10^(-7) color(white)(l) color(red)(cancel(color(black)(m))) * (10^9 color(white)(l) nm)/(1 color(white)(l) color(red)(cancel(color(black)(m))))
color(white)(lambda) = 102.6 color(white)(l) nm
References
"Bohr's Hydrogen Atom" , Chemistry Libretext