What is the value of universal gravitational constant G in units of ${g^{ - 1}}c{m^3}{s^{ - 2}}$? Given that $G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$
(A) $6.67 \times {10^{ - 8}}$
(B) $6.67 \times {10^{ - 7}}$
(C) $6.67 \times {10^{ - 9}}$
(D) $6.67 \times {10^{ - 10}}$

Hint To convert universal gravitation constant into units of ${g^{ - 1}}c{m^3}{s^{ - 2}}$
Take $N = kgm{s^{ - 2}}$
Convert meter to centimeter
Then convert kilogram to gram and put all of them in the unit $N{m^2}k{g^{ - 2}}$

Complete step-by-step answer:
According to Newton’s Law of Gravitation, the Force (F) is directly proportional to the product of their masses and is inversely proportional to square of distance between them.
$F = G\dfrac{{{m_1}{m_2}}}{{{r^2}}}$
where, ${m_1}$ and ${m_2}$ are two masses
$G = $Gravitational Constant
$r = $distance between them
To convert universal gravitational constant to ${g^{ - 1}}c{m^3}{s^{ - 2}}$ from $N{m^2}k{g^{ - 2}}$
It is given that,
$G = 6.67 \times {10^{ - 11}}N{m^2}k{g^{ - 2}}$
As we know that, $N = kgm{s^{ - 2}}$, $m = 100cm$ and $1kg = 1000g$
$\therefore G = 6.67 \times {10^{ - 11}} \times \left( {kgm{s^{ - 2}}} \right)\left( {{m^2}} \right){\left( {kg} \right)^{ - 2}}$
$G = 6.67 \times {10^{ - 11}} \times \left[ {\left( {1000g} \right) \times \left( {100cm} \right) \times {s^{ - 2}}} \right] \times {\left( {100cm} \right)^2} \times {\left( {1000g} \right)^{ - 2}}$
$G = 6.67 \times {10^{ - 11}} \times {10^3}{g^{ - 1}}c{m^3}{s^{ - 1}}$
Therefore, $G = 6.67 \times {10^{ - 8}}{g^{ - 1}}c{m^3}{s^{ - 1}}$

So, the option (A) is correct.

Note The Gravitational Constant is also known as Newtonian Constant of Gravitation and Cavendish Gravitational Constant denoted by G. It is an empirical physical constant. It is involved in the calculation of gravitation effects in Sir Isaac Newton’s law of universal gravitation and in Albert Einstein’s general theory of relativity.
The relation between $g$ and $G$ can be expressed as
$g = \dfrac{{GM}}{{{r^2}}}$