Find the longest side of the triangle.

If the side lengths are $l,\frac{2}{3}l,\frac{4}{9}l$ then the circumradius is given by: $$ R=\frac{abc}{4\Delta}=\frac{\frac{8}{27}l^3}{l^2\sqrt{\left(1+\frac{2}{3}+\frac{4}{9}\right)\left(1-\frac{2}{3}+\frac{4}{9}\right)\left(1+\frac{2}{3}-\frac{4}{9}\right)\left(-1+\frac{2}{3}+\frac{4}{9}\right)}}$$ by Heron's formula, hence: $$ R = \frac{24}{\sqrt{1463}} l $$ gives:

$$ l=\color{red}{\frac{7}{4}}.$$

Let $a,b,c$ be the side lengths. Then, $b,c$ can be written as $$b=\frac 23a,\quad c=\left(\frac 23\right)^2a$$ where $a$ is the longest side. So, we have, by the law of cosines, $$a^2=\left(\frac 23a\right)^2+\left(\left(\frac 23\right)^2a\right)^2-2\cdot\frac 23a\cdot\left(\frac 23\right)^2a\cos A$$ $$\Rightarrow \cos A=-\frac{29}{48}.$$ Hence, $$a=2R\sin A=2R\sqrt{1-\cos^2A}=2\cdot 6\sqrt{\frac{7}{209}}\sqrt{1-\left(-\frac{29}{48}\right)^2}=\frac 74.$$

The side of the triangle in G.P. with common ratio $\frac{2}{3}$ can be taken as $$a=\frac{3}{2}x, b=x, c=\frac{2}{3}x$$
Where, $x$ is a positive real number.

Hence, using cosine formula we get $$\cos C=\frac{a^2+b^2-c^2}{2ab}$$ $$=\frac{\frac{9}{4}x^2+x^2-\frac{4}{9}x^2}{2\cdot\frac{3}{2}x \cdot x }=\frac{101}{108}$$ $$\implies \sin C=\sqrt{1-\cos ^2 A}=\sqrt{1-\frac{101^2}{108^2}}=\frac{\sqrt{1463}}{108}$$ Hence, the circumscribed radius of the triangle is given by the following formula $$\frac{c}{2\sin C}$$ $$=\frac{\frac{2}{3}x}{2\frac{\sqrt{1463}}{108}}=\frac{36}{\sqrt{1463}}x$$ $$\implies \frac{36}{\sqrt{1463}}x=6\sqrt{\frac{7}{209}}$$ $$x=\frac{7}{6}$$ Now, setting the value of $x$, all the sides of the triangle are calculated as follows $$a=\frac{3}{2}x=\frac{3}{2}\cdot \frac{7}{6}=\frac{7}{4}$$ $$b=x=\frac{7}{6}$$ $$c=\frac{2}{3}x=\frac{2}{3}\cdot \frac{7}{6}=\frac{7}{9}$$

longest side is $a=\frac{7}{4}$

$$\bbox[5px, border:2px solid #C0A000]{\color{red}{\text{Longest side of the triangle},a=\color{blue}{\frac{7}{4}}}}$$