Answer
$\dfrac{3 \pi\sqrt {1+a^2}}{a^2} $
Work Step by Step
$Surface \ Area=\iint_{D} \sqrt {1+\dfrac{a^2x^2}{x^2+y^2}+\dfrac{a^2y^2}{x^2+y^2}}$
$\implies =\iint_{D} \sqrt {1+\dfrac{a^2x^2+a^2y^2}{x^2+y^2}}$
$\implies=\sqrt {1+a^2} \iint_{D} dA$
Since, $(\dfrac{1}{a})^2 \leq x^2+y^2 \leq (\dfrac{2}{a})^2 $ or, $\dfrac{1}{a^2} \leq x^2+y^2 \leq \dfrac{4}{a^2} $
Now, $Surface \ Area= \sqrt {1+a^2} \times \iint_{D} dA=\sqrt {1+a^2} \pi [(\dfrac{2}{a})^2 -\pi (\dfrac{1}{a})^2] =\dfrac{3 \pi\sqrt {1+a^2}}{a^2} $
