Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The author asks us to use Table 38.2 to calculate the de Broglie wavelength of the electron in the $n = 1, 2,$ and $3$ states of the hydrogen atom.
Recalling that
$$\lambda_n=\dfrac{h}{mv_n}$$
Hence,
$$\lambda_1=\dfrac{h}{mv_1}=\dfrac{(6.63\times 10^{-34})}{(9.11\times 10^{-31})(2.19\times 10^6)}=\color{red}{\bf 0.332 }\;\rm nm$$
$$\lambda_2=\dfrac{h}{mv_2}=\dfrac{(6.63\times 10^{-34})}{(9.11\times 10^{-31})(1.09\times 10^6)}=\color{red}{\bf 0.668}\;\rm nm$$
$$\lambda_3=\dfrac{h}{mv_3}=\dfrac{(6.63\times 10^{-34})}{(9.11\times 10^{-31})(0.73\times 10^6)}=\color{red}{\bf 0.997}\;\rm nm$$
$$\color{blue}{\bf [b]}$$
Recalling that the circumference of the orbit is given by
$$S_n=2\pi r_n$$
For $n=1$,
$$S_1=2\pi r_1=2\pi (0.053)=\color{blue}{\bf 0.333}\;\rm nm$$
which is the same length of $\boxed{S_1=\lambda_1}$
For $n=2$,
$$S_2=2\pi r_2=2\pi (0.212)=\color{blue}{\bf 1.33}\;\rm nm$$
which is twice the length of $\boxed{S_2=2\lambda_2}$
For $n=3$,
$$S_3=2\pi r_3=2\pi (0.476)=\color{blue}{\bf 2.99}\;\rm nm$$
which is triple the length of $\boxed{S_3=3\lambda_23}$
$$\color{blue}{\bf [c]}$$
For $n=3$, See the figure below.
