The identical spheres each of mass 2M are placed at the corners of...

Position vector $\overset{\to}{r}COM = \dfrac{{m}_{1}\overset{\to}{r}1 + {m}_{2}\overset{\to}{r}2 + {m}_{3}\overset{\to}{r}3}{{m}_{1} + {m}_{2} + {m}_{3}}$

\overset{\to}{r}\text{COM} = \dfrac{2M \times 0 + 2M \times 4\overset{^}{i} + 2M \times 4\overset{^}{j}}{6M}

⇒ \overset{\to}{r} = \dfrac{4}{3}\overset{^}{i} + \dfrac{4}{3}\overset{^}{j}

$| \overset{\to}{r} | = \sqrt{{\left(\dfrac{4}{3}\right)}^{2} + {\left(\dfrac{4}{3}\right)}^{2}} = \dfrac{4\sqrt{2}}{3}$

Therefore, $x = 3$.