, 2015

Solution:
$\quad \frac{\pi}{3}$

Let $B(P, r)$ be the (closed) disc centered at $P$ with radius $r$. Note that for all $(x, y) \in I$, $x > 0$, and $x > \left( \frac{y^{4}}{9} + 2015 \right)^{1/4} > \frac{|y|}{\sqrt{3}}$. Let $I' = \{ (x, y) : x \sqrt{3} > |y| \}$. Then $I \subseteq I'$ and the intersection of $I'$ with $B((0,0), r)$ is $\frac{\pi}{3} r^{2}$, so $f(r)$, the area of $I \cap B((0,0), r)$, is also less than $\frac{\pi}{3} r^{2}$. Thus $L = \frac{\pi}{3}$ works.

On the other hand, if $x > \frac{|y|}{\sqrt{3}} + 7$, then $x > \frac{|y|}{\sqrt{3}} + 7 > \left( \left( \frac{|y|}{9} \right)^{4} + 7^{4} \right)^{1/4} > \left( \frac{y^{4}}{9} + 2015 \right)^{1/4}$, which means that if $I'' = \{ (x, y) : (x-7) \sqrt{3} > |y| \}$, then $I'' \subseteq I'$. However, for $r > 7$, the area of $I'' \cap B((7,0), r-7)$ is $\frac{\pi}{3}(r-7)^{2}$, and $I'' \subseteq I$, $B((7,0), r-7) \subseteq B((0,0), r)$, which means that $f(r) > \frac{\pi}{3}(r-7)^{2}$ for all $r > 7$, from which it is not hard to see that $L = \frac{\pi}{3}$ is the minimum possible $L$.