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**Hint:**From the formula for the gravitational force between two point sized bodies, we get that $G=\dfrac{F{{r}^{2}}}{{{m}_{1}}{{m}_{2}}}$. Use this equation for G and find the SI and CGS units of G. 1N = ${{10}^{5}}$dyn, 1m = ${{10}^{2}}$cm and 1kg = ${{10}^{3}}$g, these relations will help in find the ratio of SI unit to CGS unit of G.

__Complete step-by-step answer__:G is the universal gravitational constant. It is a proportionality constant used in the equation of the gravitational force between two point sized bodies that are separated by a distance d, i.e. $F=\dfrac{G{{m}_{1}}{{m}_{2}}}{{{r}^{2}}}$

$G=\dfrac{F{{r}^{2}}}{{{m}_{1}}{{m}_{2}}}$…. (i)

Let us find the SI units of the gravitational constant G by using equation (i).

The SI unit of force F is Newton (N).

The SI unit of distance r is metre (m).

The SI units of masses ${{m}_{1}}$ and ${{m}_{2}}$ is kilogram (kg).

Therefore, the SI unit of G is $\dfrac{N{{(m)}^{2}}}{kg\times kg}=\dfrac{N{{m}^{2}}}{k{{g}^{2}}}=N{{m}^{2}}k{{g}^{-2}}$ …. (ii)

Now, let us calculate the CGS unit of G.

The CGS unit of force F is dyne (dyn).

The CGS unit of distance r is centimetre (cm).

The CGS units of masses ${{m}_{1}}$ and ${{m}_{2}}$ is gram (g).

Therefore, the CGS unit of G is $\dfrac{dyn{{(cm)}^{2}}}{g\times g}=\dfrac{(dyn)c{{m}^{2}}}{{{g}^{2}}}=(dyn)c{{m}^{2}}{{g}^{-2}}$ …. (iii)

Now divide the Si unit (ii) of G by CGS unit (iii) of G.

$\dfrac{\text{SI unit of G}}{\text{CGS unit of G}}=\dfrac{N{{m}^{2}}k{{g}^{-2}}}{(dyn)c{{m}^{2}}{{g}^{-2}}}$ …. (iv).

1N = ${{10}^{5}}$dyn

1m = ${{10}^{2}}$cm

1kg = ${{10}^{3}}$g

Substitute the values of 1N, 1m and 1kg in equation (iv).

$\Rightarrow \dfrac{\text{SI unit of G}}{\text{CGS unit of G}}=\dfrac{\left( {{10}^{5}}dyn \right){{\left( {{10}^{2}}cm \right)}^{2}}{{\left( {{10}^{3}}g \right)}^{-2}}}{(dyn)c{{m}^{2}}{{g}^{-2}}}=\dfrac{\left( {{10}^{5}}dyn \right)\left( {{10}^{4}}c{{m}^{2}} \right)\left( {{10}^{-6}}{{g}^{-2}} \right)}{(dyn)c{{m}^{2}}{{g}^{-2}}}={{10}^{3}}$

This means that the ratio of the SI unit to CGS unit of G is ${{10}^{3}}$.

**Hence, the correct option is A.**

**Note**: If you do not know the relation between the units N and dyn, then convert the N into MKS units.

We force is equal to mass times acceleration.

Therefore, the unit of force is $kgm{{s}^{-2}}$.

This means that 1N = 1$kgm{{s}^{-2}}$.

Similarly, 1dyn = 1$gcm{{s}^{-2}}$.

This means that $\dfrac{1N}{1dyn}=\dfrac{1kgm{{s}^{-2}}}{1gcm{{s}^{-2}}}=\dfrac{({{10}^{3}}g)({{10}^{2}}cm){{s}^{-2}}}{1gcm{{s}^{-2}}}={{10}^{5}}$.

Hence, 1N = ${{10}^{5}}$dyn.

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