Taking the Bohr radius as a₀ = 53 pm, the radius of Li⁺⁺ ion in its ground state, on the basis of Bohr's model, will be about

According to Bohr's model of an atom radius of an atom in its ground state is $\mathrm{r}=\frac{{\mathrm{r}}_0}{\mathrm{Z}}$
where, r₀ is Bohr's radius and Z is atomic number.
As given that,
${\mathrm{r}}_0=53$pm and atomic number of Lithium atom is 3
So, $\mathrm{r}=\frac{53}{3}=17.67\mathrm{pm}\approx 18 \mathrm{pm}$