Answer
$\rm (20\;cm,0\;cm)$
Work Step by Step
It is obvious that the $y_{cm}$ of our steel plate is zero due to the symmetry since it has the shape of the isosceles triangle.
$$y_{cm}=\color{red}{\bf 0} \;\rm cm \tag 1$$
Now we need to find its $x_{cm}$. We know that
$$x_{cm}=\int \dfrac{xdm}{M}$$
where $M$ is the whole mass of the plate.
Thus,
$$x_{cm}=\dfrac{1}{M}\int xdm\tag 2$$
As we see from the figure below, the dark-yellow rectangle has a mass of $dm$ where $dx$ is its width while $y$ is its height.
Thus, its area is given by
$$dA=y dx\tag 3$$
This area is from $[A=(y)(0)=0]$ to $[A=(20)(30)=600$ cm$^2]$.
Solving for $dx$;
And since the mass is uniformly distributed, $\sigma=\dfrac{dm}{dA}=\dfrac{M}{A}$. Solving for $dm$;
$$dm=\dfrac{MdA}{A}$$
Plugging from (3);
$$dm=\dfrac{M ydx}{A}$$
Plugging into (2);
$$x_{cm}=\dfrac{1}{M}\int x dm = \dfrac{1}{ \color{red}{\bf\not}M}\int x\dfrac{ \color{red}{\bf\not}M ydx}{A}$$
$$x_{cm}= \dfrac{1}{A}\int x ydx \tag 4$$
Now we need to find $y$; from the geometry of the figure below (the similarity between the big triangle and the small trinagle inside it), we can see that
$$\dfrac{y}{0.2}=\dfrac{x}{0.3}$$
Thus, $$y=\dfrac{2x}{3}$$
Plugging into (4);
$$x_{cm}= \dfrac{1}{A}\int x \dfrac{2x}{3}dx = \dfrac{2}{3A}\int^{0.3}_{0} x^2dx $$
$$x_{cm}= \dfrac{2}{3A} \dfrac{x^3}{3}\bigg|_0^{0.3} $$
$$x_{cm}= \dfrac{2}{3\times \frac{1}{2}(0.3)(0.2)} \dfrac{0.3^3}{3}=\color{black}{\bf 0.2} \;\rm m $$
$$x_{cm}=\color{red}{\bf 20} \;\rm cm $$