The position vector of the center of mass $$\vec{r_{cm}}$$ of an asymmetric uniform bar of negligible area of cross-section as shown in figure is:

For first segment (horizontal top bar of mass $$2m$$):

$$x_1 = L,\quad y_1 = L$$

For second segment (vertical bar of mass $$m$$):

$$x_2 = 2L,\quad y_2 = \frac{L}{2}$$

For third segment (horizontal bottom bar of mass $$m$$):

$$x_3 = \frac{5}{2}L,\quad y_3 = 0$$

Net center of mass coordinates:

$$x_{cm} = \frac{2m(L) + m(2L) + m\left(\frac{5}{2}L\right)}{2m + m + m} = \frac{13}{8}L$$

$$y_{cm} = \frac{2m(L) + m\left(\frac{L}{2}\right) + m(0)}{2m + m + m} = \frac{5}{8}L$$