Answer
$$\dfrac{3 \pi\sqrt {1+a^2}}{a^2} $$
Work Step by Step
Since, $(\dfrac{1}{a})^2 \leq x^2+y^2 \leq (\dfrac{2}{a})^2 $
$$Surface \space Area=\iint_{D} \sqrt {1+\dfrac{a^2x^2}{x^2+y^2}+\dfrac{a^2y^2}{x^2+y^2}}\\=\iint_{D} \sqrt {1+\dfrac{a^2x^2+a^2y^2}{x^2+y^2}}\\=\sqrt {1+a^2} \times\pi (\dfrac{2}{a})^2 -\pi (\dfrac{1}{a})^2 \\=\dfrac{3 \pi\sqrt {1+a^2}}{a^2} $$
