Answer
${\bf 1.5}\;\rm nm$
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the range of wavelength regions is given by:
Infrared (IR): $700 \; \rm{nm}$ to $1 \; \rm{mm}$
Visible light: $400 \;\rm{nm}$ to $700 \;\rm{nm}$
Ultraviolet (UV):$10 \; \rm{nm}$ to $400 \; \rm{nm}$
X-rays :$0.01 \; \rm{nm}$ to $ 10 \; \rm{nm}$
So the 1484-nm-photon is in the $\color{red}{\bf infrared}$ range.
$$\color{blue}{\bf [b]}$$
We know, for a particle in a one-dimensional box, that the energy levels are given by
$$
E_n = \frac{n^2 h^2}{8mL^2}\tag 1
$$
And we know that the energy difference between the $ n = 3 $ and So,
$$ E_{\rm photon}=\Delta E$$
$$\frac{hc}{\lambda} = E_3 - E_2$$
Using (1);
$$\frac{hc}{\lambda} = \frac{9h^2}{8mL^2} - \frac{4h^2}{8mL^2} = \frac{5h^2}{8mL^2} $$
Thus,
$$
\frac{ c}{\lambda}=\frac{5h}{8mL^2}
$$
Solving for $L$ to find the length of the box in which the electrons are confined.
$$
L =\sqrt{ \frac{5h\lambda}{8mc}}$$
Substituting the known;
$$
L =\sqrt{ \frac{5 (6.63 \times 10^{-34}) (1484 \times 10^{-9})}{8 (9.11 \times 10^{-31}) (3 \times 10^8)}}$$
$$L=\color{red}{\bf 1.5}\;\rm nm$$
