Cycloidish

Let the sides of the rectangle be $a$ and $b$ The area encloded can be seen as five serarate pieces.

The blue and pink together make a rectangle of area $ab$

The green and orange are quarter-circles of radius $a$ and $b$ so their areas are $\frac{1}{4}\pi a^{2}$ and $\frac{1}{4}\pi b^{2}$

The large purple quarter circle has radius $\sqrt{a^{2}+b^{2}}$ so its area is $\frac{1}{4}\pi (a^{2}+b^{2})$

Therefore the area formula for the cycloidish shape is $A= ab + \frac{1}{4}\pi a^{2}+\frac{1}{4}\pi b^{2}+\frac{1}{4}\pi (a^{2}+b^{2})$

Which simplifies to $A=\frac{\pi}{2}(a^{2}+b^{2})+ab$

Letting $a=1$ and $b=x$ , an equation to find side length that gives area $3\pi$ is

$\frac{\pi}{2}(1+x^{2})+x=3\pi$

This is a quadratic solution whose positive solution is

$x=\frac{-1+\sqrt{1+5\pi^{2}}}{\pi}\approx\boxed{1.940300568}$

Consider my diagram.

$d=\sqrt{1+x^2}$

$A_1=\dfrac{1}{4}\pi=\dfrac{\pi}{4}$

$A_2=\dfrac{1}{2}(1)(x)=\dfrac{x}{2}$

$A_3=\dfrac{1}{4}\pi (\sqrt{1+x^2})^2=\dfrac{\pi}{4}(1+x^2)=\dfrac{\pi}{4}+\dfrac{\pi x^2}{4}$

$A_4=A_2=\dfrac{x}{2}$

$A_5=\dfrac{1}{4}\pi x^2=\dfrac{\pi x^2}{4}$

The total area is $3\pi$ . So,

$3\pi = A_1+A_2+A_3+A_4+A_5$

$3\pi = \dfrac{\pi}{4}+\dfrac{x}{2}+\dfrac{\pi}{4}+\dfrac{\pi x^2}{4}+\dfrac{x}{2}+\dfrac{\pi x^2}{4}=\dfrac{2\pi x^2}{4}+\dfrac{2\pi}{4}+\dfrac{2x}{2}$

$6\pi= \pi x^2+\pi +2x$

$\pi x^2 + 2x -5\pi=0$

It ends up on a quadratic equation. By using the quadratic formula, we get

$x \approx \boxed{1.940300568}$

Very nasty numbers, probably because I'm a high school student. Basically if you can see the 3 important length and know your quadratic formula, you are good to go!

I made it on Desmos: Rolling Rectangle