Which of the following will have the smallest radius of first orbit (…

The correct answer is Option (4) → Doubly ionized lithium

In the Bohr model the radius of the nth orbit of a hydrogen-like species of atomic number Z is

    \( \displaystyle r_n = \frac{n^{2}}{Z}\,a_0, \qquad a_0 = 0.53\,\mathrm{\mathring{A}} \)

so for the first orbit \( r\propto 1/Z \): the larger the nuclear charge, the smaller the orbit.

Hydrogen (\( Z = 1 \)): \( r = 0.53\,\mathrm{\mathring{A}} \)
Deuterium (\( Z = 1 \)): \( r = 0.53\,\mathrm{\mathring{A}} \)
Singly ionized helium \( \mathrm{He^{+}} \) (\( Z = 2 \)): \( r = 0.265\,\mathrm{\mathring{A}} \)
Doubly ionized lithium \( \mathrm{Li^{2+}} \) (\( Z = 3 \)): \( r = 0.177\,\mathrm{\mathring{A}} \)

Doubly ionized lithium has the largest Z and therefore the smallest orbit.