$\sqrt {1+(\large\frac{dy}{dx})^{1/3} } =\frac{d^2 y}{dx^2}$
squaring $1+ (\large\frac{dy}{dx} )^{1/3}=( \frac{d^2 y}{dx^2})^2$
$(\large\frac{dy}{dx} )^{1/3}=( \frac{d^2 y}{dx^2})^2$$-1$
Raising both sides of the power 3.
$\large\frac{dy}{dx} = \bigg[ \bigg( \large\frac{d^2y)^2}{dx^2} \bigg)-1\bigg]^3$
Order - 2
degree -6
Hence 4 is the correct answer.