Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
In a magnetic field, an electron undergoing cyclotron motion has its angular momentum quantized according to Bohr’s condition which is given by
$$ L_n = n \hbar \tag 1$$
where the angular momentum for a charged particle moving in a circle with radius $r_n $ in a magnetic field of $B $ is given by
$$L_n = m v_n r_n\tag 2$$
From (1) and (2);
$$ n \hbar=m v_n r_n\tag 3$$
Here,$v_n $ is the speed of the electron in the $n$-th orbit. In a magnetic field.
We know that the magnetic field force exerted on the electron is given by
$$ F=\frac{m v_n^{\color{red}{\bf\not} 2}}{r_n} = e \color{red}{\bf\not} v_n B $$
Solving for$r_n $:
$$ r_n=\frac{m v_n }{e B } \tag 4$$
where, from (3), $mv_n=n\hbar /r_n$.
Hence,
$$ r_n^2=\frac{n\hbar }{e B } $$
$$ \boxed{r_n =\sqrt{\frac{n\hbar }{e B } }}$$
$$\color{blue}{\bf [b]}$$
The first four allowed radii in a 1.0 T magnetic field are for $n=1,2,3,$ and $4$.
Using the boxed formula above.
For $n=1$;
$$r_1 =\sqrt{\frac{(1)(1.055\times 10 ^{−34})}{(1.6\times 10^{-19})(1) } }=\color{red}{\bf 25.7}\;\rm nm$$
For $n=2$;
$$r_2 =\sqrt{\frac{(2)(1.055\times 10 ^{−34})}{(1.6\times 10^{-19})(1) } }=\color{red}{\bf 36.3}\;\rm nm$$
For $n=3$;
$$r_3 =\sqrt{\frac{(3)(1.055\times 10 ^{−34})}{(1.6\times 10^{-19})(1) } }=\color{red}{\bf 44.4}\;\rm nm$$
For $n=4$;
$$r_4 =\sqrt{\frac{(3)(1.055\times 10 ^{−34})}{(1.6\times 10^{-19})(1) } }=\color{red}{\bf 51.4}\;\rm nm$$
$$\color{blue}{\bf [c]}$$
We know that the cyclotron frequency is given by
$$f_{\rm cyc}=\dfrac{eB}{2\pi m}\tag 5$$
And we know that the energy associated with the circular motion (cyclotron motion) is the kinetic energy of the electron
$$E_n=\frac{1}{2}mv_n^2\tag 6$$
Solving (4) for $v_n$ and plug it into (6);
$$E_n=\frac{1}{2}m\left[ \dfrac{eBr_n}{m}\right]^2 $$
$$E_n= \dfrac{e^2B^2r_n^2}{2m} $$
Plugging $r_n$ from part (a) above;
$$E_n= \dfrac{e^{\color{red}{\bf\not} 2}B^{\color{red}{\bf\not} 2} n\hbar}{2m\color{red}{\bf\not} e\color{red}{\bf\not} B} $$
$$E_n= \dfrac{e B n\hbar}{2m } $$
\multiplying the right side by $\dfrac{\pi}{\pi}$
$$E_n= \dfrac{\pi e B n \hbar}{2\pi m } $$
$$E_n= \overbrace{ \dfrac{e B }{2\pi m }}^{\rm see \;(5)} (\pi n \hbar)$$
$$\boxed{E_n= \pi n \hbar f_{\rm cyc} }$$
