Answer
$I=\dfrac{ML^2}{6}$
Work Step by Step
We know that the moment of inertia is given by
$$I=\int r^2 dm\tag1$$
The origin point here is the geometric center of the rectangular plate.
We know that the mass is uniformly distributed, so its density is constant.
$$\rho=\dfrac{M}{V}=\dfrac{M}{At}$$
where $M$ is the plate's mass, $A$ is its surface area, and $t$ is its thickness.
Now we need to choose a small piece of this plate $dm$, as seen in the figure below as a small rectangle. The distance from this small piece $dm$ to the axis of rotation is $r$.
This $dm$ is having the same density as the plate and the same thickness but its surface area is too small and is given by $dA$.
$$\rho=\dfrac{M}{A \color{red}{\bf\not}t}=\dfrac{dm}{dA\cdot \color{red}{\bf\not} t}$$
Noting that $dA=dxdy$, and the area of the whole surface is $A=L^2$.
Hence,
$$dm=\dfrac{MdA}{A}=\dfrac{Mdxdy}{L^2}$$
Plugging into (1);
$$I=\int\int r^2 \dfrac{Mdxdy}{L^2}= \dfrac{M}{L^2}\int \int r^2dxdy$$
Noting that we have two integration here since we are moving in both directions $x$ and $y$.
We can see from the figure below that $r^2=x^2+y^2$ and the range in the $x$-direction is from $-L/2$ to $L/2$ and the range in $y$-direction is the same since the plate's surface is a square.
Thus,
$$I= \dfrac{M}{L^2}\int_{-L/2}^{L/2} \left(\int_{-L/2}^{L/2} [x^2+y^2]dx\right)dy\tag 2$$
Solving the integration between the brackets;
$$\int_{-L/2}^{L/2} [x^2+y^2]dx=\left[\dfrac{x^3}{3}+xy^2\right]_{-L/2}^{L/2}$$
$$= \dfrac{\left(\frac{L}{2}\right)^3}{3}+\left(\frac{L}{2}\right)y^2-\left[\dfrac{\left(\frac{-L}{2}\right)^3}{3}+\left(\frac{-L}{2}\right)y^2 \right]$$
$$= \dfrac{L^3}{24}+ \frac{Ly^2}{2} -\left[\dfrac{-L^3}{24}-\frac{Ly^2}{2} \right]$$
$$= \dfrac{L^3}{24}+ \frac{Ly^2}{2} + \dfrac{L^3}{24}+\frac{Ly^2}{2} = \color{blue}{\dfrac{L^3}{12}+Ly^2}$$
Plugging into (2) between the brackets.
$$I= \dfrac{M}{L^2}\int_{-L/2}^{L/2} \left(\dfrac{L^3}{12}+Ly^2\right)dy $$
$$I= \dfrac{M}{L^2 }\left(\dfrac{L^3y}{ 12}+\dfrac{Ly^3}{3}\right) \bigg|_{-L/2}^{L/2}$$
$$I= \dfrac{M}{L^2 }\left(\dfrac{L^3\left(\dfrac{L}{2}\right)}{ 12}+\dfrac{L\left(\dfrac{L}{2}\right)^3}{3}- \left[\dfrac{L^3\left(\dfrac{-L}{2}\right)}{ 12}+\dfrac{L\left(\dfrac{-L}{2}\right)^3}{3}\right]\right) $$
$$I= \dfrac{M}{L^2 }\left(\dfrac{L^4}{ 24}+\dfrac{L^4}{24}- \dfrac{-L^4}{ 24}-\dfrac{-L^4}{24}\right) =\dfrac{M}{L^2 }\left(\dfrac{4L^4}{ 24}\right) $$
$$\boxed{I=\dfrac{ML^2}{6}}$$
