Answer
$1.09707\times10^7 \;\rm m^{-1}$
Work Step by Step
As the author told us, to evaluate the Rydberg constant $R$ by using the Bohr model we will use the two formulas of
$$\dfrac{1}{\lambda}=R\left(\dfrac{1}{n'^2}-\dfrac{1}{n^2}\right)$$
and
$$\dfrac{1}{\lambda}=\dfrac{2\pi^2Z^2e^4mk^2}{h^3c}\left(\dfrac{1}{n'^2}-\dfrac{1}{n^2}\right)$$
Therefore,
$$R=\dfrac{2\pi^2Z^2e^4mk^2}{h^3c}$$
we know that $k=\dfrac{1}{4\pi \varepsilon_0}$; so
$$R=\dfrac{2\pi^2Z^2e^4m }{h^3c (4\pi \varepsilon_0)^2}=\dfrac{2\pi^2Z^2e^4m }{16h^3c \pi^2 \varepsilon_0 ^2}$$
$$R =\dfrac{ Z^2e^4m }{8 h^3c \varepsilon_0 ^2}$$
For the Hydrogen atom, $Z=1$;
$$R =\dfrac{ e^4m }{8 h^3c \varepsilon_0 ^2}$$
Plugging the known;
$$R =\dfrac{ (1.602176 \times10^{-19})^4\cdot (9.109382\times10^{-31}) }{8\cdot (6.626069 \times10^{-34})^3\cdot (2.997925\times10^8) \cdot (8.8554188 \times10^{-12})^2}$$
$$R =\color{red}{\bf 1.09707\times10^7}\;\rm m^{-1}$$
