Answer
a) ${\bf 2.95\times 10^{-10}}\;\rm m$
b) $\bf -2.44\;\rm eV$$
Work Step by Step
$$\color{blue}{\bf [a]}$$
The speed of the electron is given by
$$v=\dfrac{2\pi r}{T}$$
where $T=1/f$, so
$$v=2\pi r f$$
Recalling that $c$ which is the speed of light is given by $c=f\lambda$, so $f=c/\lambda$;
$$v=\dfrac{2\pi r c}{\lambda} \tag 1$$
The net force exerted on the electron is given by
$$\sum F_r=F_e=ma_r$$
So,
$$\dfrac{ke^2}{r^2}=\dfrac{mv^2}{r}$$
Hence,
$$v=\sqrt{\dfrac{ke^2}{mr}}\tag 2$$
Plug into (1);
$$\sqrt{\dfrac{ke^2}{mr}}=\dfrac{2\pi r c}{\lambda} $$
Squaring both sides;
$$ \dfrac{ke^2}{mr} =\dfrac{4\pi^2 r^2 c^2}{\lambda^2} $$
Solving for $r$;
$$ r^3 =\dfrac{ke^2\lambda^2}{4\pi^2 c^2m } $$
$$ r =\sqrt[3]{\dfrac{ke^2\lambda^2}{4\pi^2 c^2m } }$$
Plug the known;
$$ r =\sqrt[3]{\dfrac{(9\times 10^9)(1.6\times 10^{-19})^2(600\times 10^{-9})^2}{4\pi^2 (3\times 10^8)^2(9.11\times 10^{-31})} }$$
$$r=\color{red}{\bf 2.95\times 10^{-10}}\;\rm m$$
$$\color{blue}{\bf [b]}$$
The total energy of the hydrogen atom is the total energy of the electron (assuming that the proton of this hydrogen atom is at rest), which is the sum of its kinetic energy plus its electric potential energy.
$$E=K+U_e=\frac{1}{2}mv^2+\dfrac{k(-e)(e)}{r}$$
Thus,
$$E =\frac{1}{2}mv^2-\dfrac{k e^2}{r}\tag 1$$
Now we need to find $mv^2$ in terms of $e$ and $r$,
According to Newton's second law, the net force exerted on the electron is
$$\sum F=F_e =ma_r$$
$$\dfrac{ke^2}{r^2}=\dfrac{mv^2}{r}$$
Hence,
$$mv^2=\dfrac{ke^2}{r}\tag 2$$
Plug into (1);
$$E =\frac{1}{2}\dfrac{ke^2}{r}-\dfrac{k e^2}{r}=-\dfrac{ke^2}{2r}$$
where $k=1/4\pi \epsilon_0$
$$ E = -\dfrac{ e^2}{8\pi \epsilon_0r} $$
So, the total mechanical energy of this atom is
$$ E = \dfrac{- (1.6\times 10^{-19})^2}{8\pi (8.85\times 10^{-12})( 2.95\times 10^{-10})} $$
$$E=\color{red}{\bf -3.90\times 10^{-19}}\;\rm J=\color{red}{\bf -2.44}\;eV$$
