1. Use Bohr model expressions

For a hydrogen-like species, the total energy in the $n$th orbit is

$E_n=-\frac{13.6Z^2}{n^2}\,\text{eV}$

In Bohr's model, kinetic energy is

$K=-E_n=\frac{13.6Z^2}{n^2}\,\text{eV}$

So, the electron has the highest kinetic energy for the largest value of

$\frac{Z^2}{n^2}$

2. Evaluate each option

Option A: first orbit of $\mathrm{H}$ atom

Here, $Z=1$, $n=1$

$K \propto \frac{1^2}{1^2}=1$

Option B: first orbit of $\mathrm{He}^+$

Here, $Z=2$, $n=1$

$K \propto \frac{2^2}{1^2}=4$

Option C: second orbit of $\mathrm{He}^+$

Here, $Z=2$, $n=2$

$K \propto \frac{2^2}{2^2}=1$

Option D: second orbit of $\mathrm{Li}^{2+}$

Here, $Z=3$, $n=2$

$K \propto \frac{3^2}{2^2}=\frac{9}{4}=2.25$

3. Compare values

The proportional kinetic energies are:

$A=1,\quad B=4,\quad C=1,\quad D=2.25$

The largest value is for Option B.

4. Final answer

The highest kinetic energy is associated with the electron in the first orbit of $\mathrm{He}^+$.