Answer
See the detailed answer below.
Work Step by Step
$$\color{blue}{\bf [a]}$$
The total energy of the hydrogen atom is actually 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 $k=1/4\pi \epsilon_0$
$$\boxed{E = -\dfrac{ e^2}{8\pi \epsilon_0r}}$$
$$\color{blue}{\bf [b]}$$
$$\dfrac{U_e}{K}=\dfrac{-\dfrac{k e^2}{r}}{\dfrac{ke^2}{2r}}$$
We plugged that from (1) and (2),
$$\dfrac{U_e}{K}= \dfrac{-k e^2}{r} \dfrac{2r} {ke^2}=-2$$
Thus,
$$\boxed{U_e=-2K}$$
$$\color{blue}{\bf [c]}$$
We need to move the electron to $r=\infty$ and $E=0$, so
$$E_{\rm ionization}=|E|=13.6\;\rm eV$$
where $E$ is the total energy of the electron we found above in part a, so plugging from the first boxed formula above,
$$ \dfrac{ e^2}{8\pi \epsilon_0r}=13.6\;\rm eV$$
Solving for $r$;
$$r= \dfrac{ e^2}{8\pi \epsilon_0(13.6e)}= \dfrac{ e }{8(13.6 )\pi \epsilon_0}$$
Plug the known;
$$r= \dfrac{ (1.6\times 10^{-19})}{8(13.6 )\pi (8.85\times 10^{-12})}$$
$$r=\color{red}{\bf 5.29\times10^{-11}}\;\rm m$$
Now we need to find $v$, so we need to use the first formula above,
$$E= \frac{1}{2}mv^2-\dfrac{ke^2}{r}=-13.6 e$$
Hence,
$$ v =\sqrt{\dfrac{-13.6 e+\dfrac{ke^2}{r}}{ \frac{1}{2}m}}$$
Plug the known;
$$ v =\sqrt{\dfrac{-13.6 (1.6\times 10^{-19})+\dfrac{(9\times 10^9) (1.6\times 10^{-19})^2}{(5.29\times10^{-11})}}{ \frac{1}{2}(9.11\times 10^{-31})}}$$
$$v=\color{red}{\bf 2.19\times 10^6}\;\rm m/s$$
