It's All Limacons

For half the outer loop: $r = \sqrt{a}(1 + 2\cos(\theta))$ $(0 \leq \theta \leq \dfrac{2\pi}{3})$ .

Let $A_{1}$ be the area of the outerloop $\implies A_{1} = a\int_{0}^{\dfrac{2\pi}{3}} (1 + 2\cos(2\theta))^2 d\theta = a\int_{0}^{\dfrac{2\pi}{3}} (3 + 4\cos(\theta) + 2\cos(2\theta)) d\theta =$ $a(3\theta + 4\sin(\theta) + \sin(2\theta))|_{0}^{\dfrac{2\pi}{3}} = a(2\pi + 2\sqrt{3} - \dfrac{\sqrt{3}}{2})$ .

For half the inner loop: $r = \sqrt{a}(1 + 2\cos(\theta))$ $(\dfrac{2\pi}{3} \leq \theta \leq \pi)$ .

Let $A_{2}$ be the area of the inner loop $\implies A_{2} = a\int_{\dfrac{2\pi}{3}}^{\pi} (1 + 2\cos(2\theta))^2 d\theta = a(\pi - 2\sqrt{3} + \dfrac{\sqrt{3}}{2})$

$\implies A = A_{1} - A_{2} = a(\pi + 3\sqrt{3}) = a^2 + \dfrac{27 + 6\sqrt{3}\pi}{4} \implies$ $a^2 - (\pi + 3\sqrt{3})a + \dfrac{27 + 6\sqrt{3}\pi}{4} = 0 \implies$ $a = \dfrac{\pi + 3\sqrt{3} \pm \pi}{2} \implies a = \dfrac{2\pi + 3\sqrt{3}}{2}$ or $a = \dfrac{3\sqrt{3}}{2}$

$(0 \leq a \leq 3) \implies a = \dfrac{3\sqrt{3}}{2} \approx \boxed{2.59807621}$ .