What is the S.I. unit of the gravitational constant $G$?

Hint: The proportionality constant used in Newton’s universal law of gravitation is termed as the gravitational constant that we can write in terms of force, mass, and distance. We will figure out our desired answer from the S.I. units of the other quantities associated with Newton’s law of gravitation.

Formula used:
$F \propto \dfrac{{{m_1}.{m_2}}}{{{r^2}}}$

Complete solution:
The gravitational constant, also known as the universal gravitational constant, is an empirical constant that is involved with calculating the gravitational force of an object. According to Sir Isaac Newton’s law of universal gravitation, the gravitational force acting between two bodies is directly proportional to the product of their masses and inversely proportional to the square of the distance connecting the centers of the two bodies. Therefore we have-
$F \propto \dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
${m_1},{m_2}$ are the masses of the two bodies
$r$ is the distance between the two bodies
Hence we have-
$F = K.\dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
Where $K$ is a proportionality constant, and this constant is termed as the universal gravitational constant, denoted as $G$. Therefore, the above equation is written as-
$F = G.\dfrac{{{m_1}.{m_2}}}{{{r^2}}}$
The S.I. unit of $F$ is Newton$(N)$
The S.I. unit of ${m_1}$ and ${m_2}$ is Kilogram$(Kg)$
The S.I. unit of $r$ is meter$(m)$
From the above equation, we have-
$\Rightarrow G = F.\dfrac{{{r^2}}}{{{m_1}.{m_2}}}$
We put the S.I. units of the other quantities in the right-hand side of the equation-
$\Rightarrow G = N.\dfrac{{{{\left( m \right)}^2}}}{{{{\left( {Kg} \right)}^2}}}$
$ \Rightarrow G = \left( {Kg.m.{s^{ - 2}}} \right).\dfrac{{{{\left( m \right)}^2}}}{{{{\left( {Kg} \right)}^2}}}$
Hence, $G = {m^3}.K{g^{ - 1}}.{s^{ - 2}}$

Therefore the S.I. unit of the universal gravitational constant, i.e., $G$ is ${m^3}.K{g^{ - 1}}.{s^{ - 2}}$.

Note: The universal gravitational constant is very significant in the world of physics. The magnitude of the universal gravitational constant is $6.674 \times {10^{ - 11}}{m^3}.K{g^{ - 1}}.{s^{ - 2}}$. It is also useful in Einstein’s General Theory of Relativity. In Einstein’s field equations, this constant quantifies the relation between the energy-momentum tensor and space-time geometry.