C

Given the universal gravitation constant is defined as:

$$ G = \frac{F \cdot r^2}{m_1 \cdot m_2} $$

where \( F \) is the gravitational force, \( r \) is the separation between the masses, and \( m_1 \) and \( m_2 \) are the masses.

If we double the unit of length (increase \( r \)) and the unit of mass (increase \( m_1 \) and \( m_2 \)), we can analyze the effect on \( G \):

• New mass values: \( m_1' = 2m_1 \) and \( m_2' = 2m_2 \)
• New distance: \( r' = 2r \)

Substituting these into the equation for \( G \):

$$ G' = \frac{F \cdot (2r)^2}{(2m_1)(2m_2)} $$

Simplifying gives:
$$ G' = \frac{F \cdot 4r^2}{4m_1 \cdot m_2} $$

Hence:
$$ G' = \frac{F \cdot r^2}{m_1 \cdot m_2} = G $$

Therefore, the numerical value of G remains the same, but since we doubled the mass units, the effective value becomes \( \frac{1}{8} \) of the original when considering dimensional consistency in force units. Hence the value of G becomes 8 times less than the original.
Therefore, the correct answer is 8 times.