According to Bohr's model, the highest kinetic energy is associated with the electron in the

For any one-electron (hydrogen-like) species, Bohr’s model gives the total energy in the $$n^{\text{th}}$$ orbit as
$$E_n = -\frac{13.6\,\text{eV}\;Z^{2}}{n^{2}}$$
where $$Z$$ is the atomic (nuclear) charge.

The kinetic energy in that orbit is related to the total energy by
$$KE_n = -E_n$$
(because in Bohr’s model $$E_n = KE_n + PE_n$$ with $$PE_n = -2\,KE_n$$).

Hence
$$KE_n = \frac{13.6\,\text{eV}\;Z^{2}}{n^{2}} \quad -(1)$$
so $$KE_n$$ is directly proportional to $$\dfrac{Z^{2}}{n^{2}}$$.

Now evaluate $$\dfrac{Z^{2}}{n^{2}}$$ for each option:

Option A: $$H$$ atom  ($$Z = 1,\; n = 1$$)
$$\dfrac{Z^{2}}{n^{2}} = \dfrac{1^{2}}{1^{2}} = 1$$

Option B: $$He^{+}$$  ($$Z = 2,\; n = 1$$)
$$\dfrac{Z^{2}}{n^{2}} = \dfrac{2^{2}}{1^{2}} = 4$$

Option C: $$He^{+}$$  ($$Z = 2,\; n = 2$$)
$$\dfrac{Z^{2}}{n^{2}} = \dfrac{2^{2}}{2^{2}} = 1$$

Option D: $$Li^{2+}$$  ($$Z = 3,\; n = 2$$)
$$\dfrac{Z^{2}}{n^{2}} = \dfrac{3^{2}}{2^{2}} = \dfrac{9}{4} = 2.25$$

The largest value is $$4$$ for Option B, meaning the electron in the first orbit of $$He^{+}$$ possesses the highest kinetic energy.

Therefore, the correct choice is:
Option B which is: First orbit of $$He^{+}$$