Answer
a) $\lambda_{2 \rightarrow 1} = \dfrac{8mL^2 c}{3h}$
b) $0.795\;\rm nm$
Work Step by Step
$$\color{blue}{\bf [a]}$$
In a rigid box (also called an infinite potential well), the energy levels of a particle are given by
$$
E_n = \frac{n^2 h^2}{8mL^2}
$$
So for a transition from $ n = 2 $ to $ n = 1 $
$$
\Delta E = E_2 - E_1 = \frac{4h^2}{8mL^2} - \frac{h^2}{8mL^2} = \frac{3h^2}{8mL^2}
$$
Recalling that the energy of the emitted photon is equal to the energy difference, so
$$
\Delta E=E_{\rm photon}
$$
Thus,
$$
\frac{3h^2}{8mL^2} = \frac{hc}{\lambda_{2 \rightarrow 1}}
$$
Solve for $ \lambda_{2 \rightarrow 1} $:
$$
\boxed{\lambda_{2 \rightarrow 1} = \frac{8mL^2 c}{3h}}
$$
$$\color{blue}{\bf [b]}$$
To find the length of the box for a given wavelength of 694 nm, we need to solve the boxed formula above for $L$.
So,
$$
L^2 = \frac{3h \lambda_{2 \rightarrow 1}}{8mc}
$$
$$
L = \sqrt{\frac{3h \lambda_{2 \rightarrow 1}}{8mc}}
$$
Substitute the known;
$$
L = \sqrt{\frac{3( 6.63 \times 10^{-34}) (694 \times 10^{-9} )}{8(9.11 \times 10^{-31})(3.0 \times 10^8)}}=\bf 7.95\times 10^{-10}\;\rm m
$$
$$L=\color{red}{\bf 0.795}\;\rm nm$$
