We know that K.E. and P.E of the electron in the $$n^{th}$$ energy state of hydrogen atom are respectively given by,
$$E_{k}=\dfrac{1}{4\pi \epsilon}\times\dfrac{e^{2}}{2r}$$
and $$E_{p}=-\dfrac{1}{4\pi \epsilon _{0}}\times \dfrac{e^{2}}{r_{n}}$$
Where, $$r_{n}=4\pi \epsilon _{0}\times\dfrac{n^{2}h^{2}}{4\pi ^{2}me^{2}}$$
The first excited state corresponds to $$n=2$$, level. Now, total energy of electron in $$n=2$$, state is given to be $$-3.4eV$$.
$$\therefore \dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{2r_{2}}+\left ( -\dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{r_{2}} \right )=-3.4eV$$
$$\Rightarrow \dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{2r_{2}}=3.4eV$$
(a) Therefore, K.E of the electron in the first excited state $$(n=2)$$ of hydrogen atom,
$$=\dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{2r_{2}}=3.4eV$$.
(b) $$E$$ of the electron in the first excited state $$(n=2)$$ of the hydrogen atom,
$$=\dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{2r_{2}}=-2\times\left ( \dfrac{1}{4\pi \epsilon _{0}}\times\dfrac{e^{2}}{2r_{2}} \right )$$
$$=-2\times3.4$$
$$=-6.8eV$$.
(c) If the zero of potential energy is chosen differently, K.E does not change. The P.E and total energy will change.