Kepler’s third law states that square of period of revolution $$\left( T \right)$$ of a planet around the sun, is proportional to third power of average distance $$r$$ between the sun and planet i.e. $${T^2} = K{r^3},$$   here $$K$$ is constant. If the masses of the sun and planet are $$M$$ and $$m$$ respectively, then as per Newton’s law of gravitation force of attraction between them is
$$F = \frac{{GMm}}{{{r^2}}},$$   here $$G$$ is gravitational constant. The relation between $$G$$ and $$K$$ is described as

Solution :
The gravitational force of attraction between the planet and sun provide the centripetal force
i.e. $$\frac{{GMm}}{{{r^2}}} = \frac{{m{v^2}}}{r}$$
$$ \Rightarrow v = \sqrt {\frac{{GM}}{r}} $$
The time period of planet will be
$$T = \frac{{2\pi r}}{v} \Rightarrow {T^2} = \frac{{4{\pi ^2}{r^2}}}{{\frac{{GM}}{r}}} = \frac{{4{\pi ^2}{r^3}}}{{GM}}\,......\left( {\text{i}} \right)$$
Also from Kepler's third law
$${T^2} = K{r^3}\,......\left( {{\text{ii}}} \right)$$
From Eqs. (i) and (ii), we get
$$\frac{{4{\pi ^2}{r^3}}}{{GM}} = K{r^3} \Rightarrow GMK = 4{\pi ^2}$$