Three rods of same mass are placed as shown in figure. The co-ordinates of centre of mass of the system are

The center of mass coordinates for the system of three uniform rods, each of mass $M$, are determined by locating each rod's center and applying the weighted average formula.

The vertical rod from $(0,0)$ to $(0,a)$ has center at $\left(0, \frac{a}{2}\right)$.

The horizontal rod from $(0,0)$ to $(2a,0)$ has center at $(a, 0)$.

The hypotenuse rod from $(0,a)$ to $(2a,0)$ has center at $\left(a, \frac{a}{2}\right)$.

With total mass $3M$, the system's center of mass coordinates are:

$X_{CM} = \frac{M(0) + M(a) + M(a)}{3M} = \frac{2a}{3}$

$Y_{CM} = \frac{M\left(\frac{a}{2}\right) + M(0) + M\left(\frac{a}{2}\right)}{3M} = \frac{a}{3}$

The result $\left(\frac{2a}{3}, \frac{a}{3}\right)$ corresponds to option D.

Final answer: $\boxed{\text{D}}$