Due to gravitational attraction, the earth revolves round the sun....

According to Bohr's quantization rule, the angular momentum of the earth is quantized as:

$M_e v r = \dfrac{nh}{2\pi}$

The gravitational force provides the necessary centripetal force for the earth's orbit:

$\dfrac{G M_s M_e}{r^2} = \dfrac{M_e v^2}{r}$

$\Rightarrow v = \sqrt{\dfrac{G M_s}{r}}$

Substituting the expression for $v$ into the quantization rule:

$M_e \sqrt{\dfrac{G M_s}{r}} r = \dfrac{nh}{2\pi}$

$M_e \sqrt{G M_s r} = \dfrac{nh}{2\pi}$

Squaring both sides, we get the radius of the $n$-th orbit:

$r = \dfrac{n^2 h^2}{4\pi^2 G M_s M_e^2}$

For the minimum radius, $n = 1$. Substituting the standard values ($h = 6.63 \times 10^{-34} \text{ J s}$, $G = 6.67 \times 10^{-11} \text{ N m}^2/\text{kg}^2$) and the given masses ($M_s = 2.0 \times 10^{30} \text{ kg}$, $M_e = 6.0 \times 10^{24} \text{ kg}$):

$r = \dfrac{(1)^2 \times (6.63 \times 10^{-34})^2}{4 \pi^2 \times (6.67 \times 10^{-11}) \times (2.0 \times 10^{30}) \times (6.0 \times 10^{24})^2}$

$r = \dfrac{43.96 \times 10^{-68}}{4 \times 9.87 \times 6.67 \times 10^{-11} \times 2.0 \times 10^{30} \times 36.0 \times 10^{48}}$

$r = \dfrac{43.96 \times 10^{-68}}{18958 \times 10^{67}}$

$r \approx 2.3 \times 10^{-138} \text{ m}$