Let $a=$ side length of equilateral triangle, $r=$ radius of circle and $x=$ resistance per unit length of wire used. Then, $L=3 a=2 \pi r$ or $a=\frac{L}{3}$ and $r=\frac{L}{2 \pi}$ Now,

Equivalent resistance across $A B$ is

$R_{A B}=(a x \| 2 a x)=\frac{a x \times 2 a x}{a x+2 a x}$

$=\frac{2 a^2 x^2}{3 a x}=\frac{2}{3} a x$

$\Rightarrow \quad R_{A B}=\frac{2}{3} \times \frac{L}{3} \times x$

Power dissipated is

$P_1=\frac{V^2}{R_{A B}}=\frac{9 V^2}{2 L x} \quad \dots(i)$

$R_{P Q}=(\pi r x \| \pi r x)=\frac{\pi r x \times \pi r x}{\pi r x+\pi r x}=\frac{\pi^2 r^2 x^2}{2 \pi r x}$

$=\frac{1}{2} \pi r x=\frac{1}{2} \pi \times \frac{L}{2 \pi} x=\frac{L x}{4}$

So, power dissipated is

$P_2=\frac{V^2}{R_{P Q}}=\frac{4 V^2}{L x}$

Ratio of power dissipated in two cases is

$\frac{P_1}{P_2}=\frac{9 V^2 / 2 L x}{4 V^2 / L x}=\frac{9}{8}$