According to Bohr's model of the hydrogen atom, the kinetic energy of an electron in a particular orbit is given by the expression:

$$ K.E. = \frac{1}{2}mv^2 $$

Where $m$ is the mass of the electron and $v$ is its velocity. However, a more useful expression involving only constants is:

$$ K.E. = \frac{13.6 \, \text{eV} \, Z^2}{n^2} $$

Where $Z$ is the atomic number and $n$ is the principal quantum number of the orbit. This formula shows that the kinetic energy of the electron is directly proportional to the square of the atomic number $Z$ and inversely proportional to the square of the orbit number $n$.

Let's analyze each option keeping the above formula in mind:

Option A: First orbit of hydrogen ($\mathrm{H}$, $Z = 1$, $n = 1$)

$$ K.E. = \frac{13.6 \, \text{eV} \, (1)^2}{(1)^2} = 13.6 \, \text{eV} $$

Option B: First orbit of $ \mathrm{He}^{+} $ ($Z = 2$, $n = 1$)

$$ K.E. = \frac{13.6 \, \text{eV} \, (2)^2}{(1)^2} = 54.4 \, \text{eV} $$

Option C: Second orbit of $ \mathrm{He}^{+} $ ($Z = 2$, $n = 2$)

$$ K.E. = \frac{13.6 \, \text{eV} \, (2)^2}{(2)^2} = 13.6 \, \text{eV} $$

Option D: Second orbit of $ \mathrm{Li}^{2+} $ ($Z = 3$, $n = 2$)

$$ K.E. = \frac{13.6 \, \text{eV} \, (3)^2}{(2)^2} = 30.6 \, \text{eV} $$

From the above calculations, it is evident that the highest kinetic energy is associated with the electron in the first orbit of $ \mathrm{He}^{+} $ (Option B), which is $54.4 \, \text{eV}$.