A rod of length 5L is bent right angle keeping one side length as 2L.

The position of the centre of mass of the system: (Consider L = 10 cm)

The bent rod can be treated as two uniform rods of lengths $$3L$$ and $$2L$$ joined at the corner.

For the horizontal rod, its centre of mass is at  $$x_1=L,\qquad y_1=0$$

For the vertical rod, its centre of mass is at  $$x_2=0,\qquad y_2=\frac{3L}{2}$$

Since the rod is uniform, mass is proportional to length.

Thus,  $$x_{cm}=\frac{(2L)(L)+(3L)(0)}{5L}$$

$$x_{cm}=\frac{2L}{5}$$

Similarly,  $$y_{cm}=\frac{(2L)(0)+(3L)\left(\frac{3L}{2}\right)}{5L}$$

$$y_{cm}=\frac{9L}{10}$$

Given,  $$L=10\ \mathrm{cm}$$

Therefore,  $$x_{cm}=4\ \mathrm{cm}$$

$$y_{cm}=9\ \mathrm{cm}$$

Hence, the position vector of the centre of mass is  $${4\hat{i}+9\hat{j}\ \mathrm{cm}}$$

Hence, the correct option is D.