$\text{ Let }{G}_1,G_2,G_3\text{ be the centroids of the triangular faces }\mathrm{OBC},\mathrm{OCA},\mathrm{OAB}$$\text{ of a tetrahedron }\mathrm{OABC}\text{ . If }{V}_1\text{ denote the volume of the tetrahedron }\mathrm{OABC}\text{ and }{V}_2\text{ that of }$$\text{ the parallelopiped with }O{G}_1,O{G}_2,O{G}_3\text{ as three }$$\text{ concurrent edges, then : }$

$\text{ Taking }O\text{ as the origin,let the position vectors of }A,B\text{ and }$$\mathrm{C}\text{ be }\overrightarrow{a},\overrightarrow{b}\text{ and }\overrightarrow{c}\text{ respectively. Then the position vectors }$of

$G_1,G_2\text{ and }{G}_3\text{ are }\frac{\overrightarrow{b}+\overrightarrow{c}}{3},\frac{\overrightarrow{c}+\overrightarrow{a}}{3}\text{ and }\frac{\overrightarrow{a}+\overrightarrow{b}}{3}\text{ respectively. }$

$\therefore V_1=\frac{1}{6}\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$

$\text{ and }{V}_2=\left[O\overrightarrow{G_1}O\overrightarrow{G_2}O\overrightarrow{G_3}\right]$

$\text{ Now, }V2=\left[O\overrightarrow{G_1}O\overrightarrow{G_2}O\overrightarrow{G_3}\right]$

$\Rightarrow V2=\left[\frac{\overrightarrow{b}+\overrightarrow{c}}{3}\frac{\overrightarrow{c}+\overrightarrow{a}}{3}\frac{\overrightarrow{a}+\overrightarrow{b}}{3}\right]$

$\Rightarrow V2=\frac{1}{27}\left[\overrightarrow{b}+\overrightarrow{c}\overrightarrow{c}+\overrightarrow{a}\overrightarrow{a}+\overrightarrow{b}\right]$

$\Rightarrow V2=\frac{2}{27}\left[\overrightarrow{a}\overrightarrow{b}\overrightarrow{c}\right]$

$\Rightarrow V2=\frac{2}{27}\text{\hspace{0.17em}×\hspace{0.17em}}6V_1\Rightarrow 9V_2=4V_1$

$\text{ Hence, }\text{ (a) is the correct answer. }$