Answer
$ 0.0373\;\rm kg\cdot m^2$
Work Step by Step
We know that the moment of inertia is given by
$$I=\int r^2 dm\tag 1$$
We know that the mass is uniformly distributed, so the plate's 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 $dA=dxdy$.
$$\rho =\dfrac{M}{A\color{red}{\bf\not} t}=\dfrac{dm}{dA\cdot\color{red}{\bf\not} t}$$
Hence,
$$dm=\dfrac{MdA}{A}=\dfrac{Mdxdy}{A}$$
Plugging into (1); (Noting that we are moving in both directions so we must have two integrations, as seen below);
$$I=\int_{y=-0.1}^{y=0.1}\int_{x=0}^{x=0.3} r^2 \dfrac{Mdxdy}{A}$$
$$I=\dfrac{M}{A}\int_{y=-0.1}^{y=0.1}\int_{x=0}^{x=0.3} r^2 dxdy$$
From the geometry in the figure below, we can see that $r^2=x^2+y^2$
$$I=\dfrac{M}{A}\int_{x=0}^{x=0.3}\left[ \int_{y=-0.1}^{y=0.1}(x^2+y^2) dy\right]dx $$
We can see that the upper part of the given plate has a slope of $\frac{1}{3}$ while the lower part has a slope of $\frac{-1}{3}$. Thus, for the upper part $y=\frac{1}{3}x$ and the for the lower part $y=\frac{-1}{3}x$. Hence $\frac{-1}{3}x\leq y\leq \frac{1}{3}x$;
$$I=\dfrac{M}{A}\int_{x=0}^{x=0.3}\left[ \color{blue}{ \int_{\frac{-1}{3}x}^{\frac{ 1}{3}x}(x^2+y^2) dy}\right]dx \tag 2$$
Solving the integration between the brackets;
$$\int_{\frac{-1}{3}x}^{\frac{ 1}{3}x} (x^2+y^2) dy=x^2y+\dfrac{y^3}{3}\bigg|_{\frac{-1}{3}x}^{\frac{ 1}{3}x} $$
$$=x^2\left[\frac{ 1}{3}x\right]+\dfrac{\left[\frac{ 1}{3}x\right]^3}{3}-\left( x^2\left[\frac{- 1}{3}x\right]+\dfrac{\left[\frac{ -1}{3}x\right]^3}{3}\right)$$
$$=\dfrac{x^3}{3}+\dfrac{x^3}{81}+\dfrac{x^3}{3}+\dfrac{x^3}{81}=\color{blue}{\dfrac{56x^3}{81}}$$
Plugging into (2);
$$I=\dfrac{M}{A}\int_{x=0}^{x=0.3}\dfrac{56x^3}{81} dx=\dfrac{56M}{81A}\int_{x=0}^{x=0.3} x^3 dx$$
$$I=\dfrac{56M}{81A}\;\dfrac{x^4}{4}\bigg|_{x=0}^{x=0.3} =\dfrac{14M}{81A}\; x^4 \bigg|_{x=0}^{x=0.3} $$
$$I=\dfrac{14M}{81A}\; (0.3^4-0^4)=\dfrac{0.3^4\times 14\times 0.8}{81\times \frac{1}{2}\times 0.3\times 0.2}$$
$$I=\color{red}{\bf 0.0373}\;\rm kg\cdot m^2$$
