Answer
a) $-1.5065\times 10^{-20}\;\rm J$
b) $6.365\times 10^{-10}\;\rm m$
Work Step by Step
We know that
$$F=\dfrac{ke^2}{r^2}$$
where $k=1/4\pi \varepsilon_0$;
$$F=\dfrac{ e^2}{4\pi \varepsilon_0 r^2}$$
and the author told us to substitute $( \varepsilon_0 =K\varepsilon_0 )$.
Thus,
$$F=\dfrac{ e^2}{4\pi K\varepsilon_0 r^2} $$
Now we can see that
$$\dfrac{ e^2}{\color{red}{\bf\not}4\color{red}{\bf\not}\pi \color{red}{\bf\not}\varepsilon_0 \color{red}{\bf\not}r^2}=\dfrac{ e^2}{\color{red}{\bf\not}4\color{red}{\bf\not}\pi K\color{red}{\bf\not}\varepsilon_0 \color{red}{\bf\not}r^2}$$
$$e^2=\dfrac{e^2}{K}\tag 1$$
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a) The energy of the electron is given by
$$E_n=\dfrac{-2\pi^2Z^2e^4m_ek^2}{h^2n^2}$$
and since the electron is in the ground state, thus $n=1$;
$$E_1=\dfrac{-2\pi^2Z^2e^4m_ek^2}{h^2 }$$
And since the arsenic ion has a $+1$ charge, thus $Z=1$ since it loses one electron to have this amount of charge.
$$E_1=\dfrac{-2\pi^2 m_ek^2}{h^2 }\;e^4$$
Plugging from (1);
$$E_1=\dfrac{-2\pi^2 m_ek^2}{h^2 }\;\dfrac{e^4}{K^2}$$
Plugging the known;
$$E_1=\dfrac{-2\pi^2 (9.11\times 10^{-31})(8.99\times 10^9)^2}{(6.626\times 10^{-34})^2 }\;\dfrac{(1.6\times 10^{-19})^4}{12^2}$$
$$E_1=\color{red}{\bf -1.5065\times 10^{-20}}\;\rm J=\color{red}{\bf -0.09416}\;\rm eV$$
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b) The radius of the electron orbit is given by
$$r_n=\dfrac{n^2h^2}{4\pi^2 m_ekZe^2}$$
for the ground state $n=1$, and for the arsenic ion $Z=1$;
$$r_1=\dfrac{ h^2}{4\pi^2 m_ek e^2}$$
Plugging from (1);
$$r_1=\dfrac{ Kh^2}{4\pi^2 m_ek e^2}$$
Plugging the known;
$$r_1=\dfrac{ 12(6.626\times 10^{-34})^2}{4\pi^2 (9.11\times 10^{-31}) (8.99\times 10^9) (1.6\times 10^{-19})^2}$$
$$r_1=\color{red}{\bf 6.365\times 10^{-10}}\;\rm m$$