Answer
$$
\begin{aligned}
I_{x}& =\frac{1}{3} \rho b h^{3} ,
\end{aligned}
$$
$$
\begin{aligned}
I_{y} &=\frac{1}{3} \rho b^{3} h,
\end{aligned}
$$
$$
\overline{\bar{y}}=\frac{h}{\sqrt{3}},
$$
$$
\overline{\bar{x}}=\frac{b}{\sqrt{3}}.
$$
Work Step by Step
A lamina with constant density $\rho(x, y)= \rho $ occupies the given region :
The rectangle is: $0 \leq x \leq b,\quad 0 \leq y \leq h$.
The moment of inertia of the lamina about the $x$-axis is:
$$
\begin{aligned}
I_{x}& =\iint_{D} y^{2} \rho(x, y) d A \\
&=\int_{0}^{h} \int_{0}^{b} \rho y^{2} d x d y \\
&=\rho \int_{0}^{b} d x \int_{0}^{h} y^{2} d y \\
&=\rho[x]_{0}^{b}\left[\frac{1}{3} y^{3}\right]_{0}^{h}\\
&=\rho b\left(\frac{1}{3} h^{3}\right) \\
&=\frac{1}{3} \rho b h^{3}
\end{aligned}
$$
and the moment of inertia of the lamina about the $y$-axis is:
$$
\begin{aligned}
I_{y} &=\iint_{D} x^{2} \rho(x, y) d A \\
&=\int_{0}^{h} \int_{0}^{b} \rho x^{2} d x d y \\
&=\rho \int_{0}^{b} x^{2} d x \int_{0}^{h} d y \\
&=\rho\left[\frac{1}{3} x^{3}\right]_{0}^{b}[y]_{0}^{h} \\
&=\frac{1}{3} \rho b^{3} h.
\end{aligned}
$$
Notice that the mass of the rectangle is
$m =\text { density * area }= \rho* bh $.
So, the radius of gyration $\overline{\bar{y}}$ with respect to the $x$-axis is given by:
$$
\overline{\bar{y}}^{2}=\frac{I_{x}}{m}=\frac{\frac{1}{3} \rho b h^{3}}{\rho b h}=\frac{h^{2}}{3} \Rightarrow \overline{\bar{y}}=\frac{h}{\sqrt{3}}
$$
and the radius of gyration $\overline{\bar{x}}$ with respect to the $y$-axis is given by:
$$
\overline{\bar{x}} ^{2}=\frac{I_{y}}{m}=\frac{\frac{1}{3} \rho^{3} h}{\rho b h}=\frac{b^{2}}{3} \Rightarrow \overline{\bar{x}}=\frac{b}{\sqrt{3}}
$$
