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

According to Bohr's quantization rule, the angular momentum of the earth is given by:

$L = \dfrac{nh}{2\pi}$

The angular momentum of the earth revolving in a circular orbit is:

$L = mvr$

The orbital speed $v$ of the earth is obtained by equating the gravitational force to the centripetal force:

$\dfrac{GMm}{r^2} = \dfrac{mv^2}{r} \Rightarrow v = \sqrt{\dfrac{GM}{r}}$

Substituting $v$ into the expression for angular momentum:

$L = m \sqrt{\dfrac{GM}{r}} r = m\sqrt{GMr}$

Equating the two expressions for angular momentum:

$\dfrac{nh}{2\pi} = m\sqrt{GMr}$

$n = \dfrac{2\pi m}{h} \sqrt{GMr}$

Substituting the given values ($G = 6.67 \times 10^{-11} \text{ N m}^2 \text{ kg}^{-2}$, $h = 6.63 \times 10^{-34} \text{ J s}$):

$n = \dfrac{2 \times 3.14 \times 6.0 \times 10^{24}}{6.63 \times 10^{-34}} \sqrt{6.67 \times 10^{-11} \times 2.0 \times 10^{30} \times 1.5 \times 10^{11}}$

$n = \dfrac{37.68 \times 10^{24}}{6.63 \times 10^{-34}} \sqrt{20.01 \times 10^{30}}$

$n \approx 5.68 \times 10^{58} \times 4.47 \times 10^{15}$

$n \approx 25.4 \times 10^{73} = 2.54 \times 10^{74}$

The value of the principal quantum number $n$ is approximately $2.5 \times 10^{74}$.