Answer
a) $T_n = {\bf 1.52\times 10^{-16}}\;n^3 $
b) ${\bf 1.32 \times 10^6}\;\rm rev $
Work Step by Step
$$\color{blue}{\bf [a]}$$
We know that the orbital speed of an electron is given by
$$v=\dfrac{2\pi r}{T}$$
So the orbital period in the $n$th is given by
$$T_n=\dfrac{2\pi r_n}{v_n}$$
Recalling that $r_n=n^2a_B$, and $v_n=v_1/n$.
Thus,
$$T_n=\dfrac{2\pi n^3a_B}{v_1}=\dfrac{2\pi a_B}{v_1}n^3$$
where $a_B=r_1$
$$T_n =\overbrace{ \left[\dfrac{2\pi r_1}{v_1}\right]}^{T_1} n^3$$
$$\boxed{T_n = n^3T_1}$$
The numerical value of $T_1$;
$$T_1=\dfrac{2\pi (0.0529\times 10^{-9})}{(2.19\times 10^6)}=
\color{red}{\bf 1.52\times 10^{-16}}\;\rm s$$
Therefore,
$$\boxed{T_n = \color{red}{\bf 1.52\times 10^{-16}}\;n^3 }$$
$$\color{blue}{\bf [b]}$$
The number of revolutions before the electron makes a quantum jump to the $n =1$ state is given by
$$N=\dfrac{\Delta t}{T_2}$$
where $\Delta t $ is the time the electron states in $n=2$ state before undergoing a quantum jump.
Plug the known and $T_2$ from the boxed formula above.
$$N=\dfrac{(1.6\times 10^{-9})}{(1.52\times 10^{-16})n^3}$$
where $n=2$;
$$N=\dfrac{(1.6\times 10^{-9})}{(1.52\times 10^{-16})(2)^3}= \color{red}{\bf 1.32 \times 10^6}\;\rm rev $$
