Centre of mass of the uniform rod will lie at its centre
\(x_{ cm }=\frac{m_1 x_1+m_2 x_2}{m_1+m_2}\)
\(x_{ cm }=\frac{m\left(\frac{L}{2}\right)+2 m(0)}{3 m}, \quad y_{ cm }=\frac{2 m(L)+m(0)}{3 m}\)
\(x_{ cm }=\frac{L}{6}, \quad y_{ cm }=\frac{2 L}{3}\)