Do you need a hint?

Consider the diagram. We can see that

$3a+2y=30$ or $3a=30-2y$

and

$3a+2x=15$ or $3a=15-2x$

Since $3a=3a$ , we have

$30-2y=15-2x$

$x=\dfrac{2y-15}{2}$ $\implies$ $\color{#3D99F6}\boxed{1}$

The area of the four blue rectangles is $30(15)-234=216$ . So the area of one blue rectangle is

$xy=\dfrac{216}{4}=54$ $\implies$ $\color{#3D99F6}\boxed{2}$

Now substitute $\color{#3D99F6}\boxed{1}$ in $\color{#3D99F6}\boxed{2}$ , we have $\left(\dfrac{2y-15}{2}\right)(y)=54$

$2y^2-15y=108$

By using the quadratic formula, we get

$\boxed{y=12}$

Let x = the thickness of yellow area.

$(30 * 15)-(30-3x)(15-3x)=234$

then x = 2 or 13 (13 will be neglected as the dimensions will be less than 0)

the required length $= (30 - 3x)/2 = (30-6)/2 = \boxed{12}$