Answer
$\lambda = 51.2~nm$
Work Step by Step
We can write a general equation for the energy:
$E_n = \frac{n^2~h^2}{8mL^2}$
We can find the energy when $n = 1$:
$E_1 = \frac{(1)^2~(6.626\times 10^{-34}~J~s)^2}{(8)(207)(9.109\times 10^{-31}~kg)(15\times 10^{-12}~m)^2}$
$E_1 = 1.294\times 10^{-18}~J$
We can find the energy when $n = 2$:
$E_2 = \frac{(2)^2~(6.626\times 10^{-34}~J~s)^2}{(8)(207)(9.109\times 10^{-31}~kg)(15\times 10^{-12}~m)^2}$
$E_2 = 5.174\times 10^{-18}~J$
We can find the energy difference:
$\Delta E = E_2 - E_1$
$\Delta E = (5.174\times 10^{-18}~J) - (1.294\times 10^{-18}~J)$
$\Delta E = 3.88\times 10^{-18}~J$
We can find the wavelength of the photon that is emitted:
$\Delta E = \frac{h~c}{\lambda}$
$\lambda = \frac{h~c}{\Delta E}$
$\lambda = \frac{(6.626\times 10^{-34}~J~s)(3.0\times 10^8~m/s)}{3.88\times 10^{-18}~J}$
$\lambda= 5.12\times 10^{-8}~m$
$\lambda = 51.2~nm$
