Explanation

(c) Let the mass of an element of length $dx$ of the rod located at a distance $x$ away from left end is $\frac{M}{L}dx$. The $x$- coordinate of the centre of mass is given by
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Total mass of rod $= \int_{0}^{L}{Ax\ dx = \frac{AL^{2}}{2}}$
$X_{cm} = \frac{1}{M}\int_{}^{}{x\ dm = \frac{1}{\left( \frac{AL^{2}}{2} \right)}\int_{0}^{L}{x\ (Ax\ dx)}}$
$= \frac{2A}{AL^{2}}\left\lbrack \frac{x^{3}}{3} \right\rbrack_{0}^{L} = \left\lbrack \frac{2}{L^{2}} \right\rbrack\left\lbrack \frac{L^{3}}{3} \right\rbrack = \frac{2L}{3}$
Hence, the centre of mass is at $\left( \frac{2L}{3},0,\ 0 \right)$