Three identical spheres, each of mass M, are placed at the corners...

First, recall the formula of the center of mass of $n$ particle system,

${r}_{cm} = \dfrac{{m}_{1}{r}_{1} + {m}_{2}{r}_{2} + {m}_{3}{r}_{3} + . . . + {m}_{n}{r}_{n}}{{m}_{1} + {m}_{2} + {m}_{3} + . . + {m}_{n}}$

$⇒ {x}_{com} = \dfrac{M \times 0 + M \times 2 + M \times 0}{3M} = \dfrac{2}{3}$

$⇒ {y}_{\text{com}} = \dfrac{M \times 0 + M + 2 + M \times 0}{3M} = \dfrac{2}{3}$

So the position vector of center of mass is, \left(\dfrac{2}{3}\overset{^}{i} + \dfrac{2}{3}\overset{^}{j}\right).