Solvig $x=2$ & $ y=2x$
We have $y=4$
$OA= 2\qquad AB= 4$
$OB =\sqrt {4+ 16} =2 \sqrt {5}$
Required surface area = surface area of the cone of radius 4 height 2.
$\qquad= \pi rl=\pi \times 4 \times 2 \sqrt 5$
$\qquad= 8 \sqrt 5\; \pi$
Hence 1 is the correct answer.