Answer
$c=\frac{3k}{2d}$
Work Step by Step
Given that the $F= -Kx-cx^{2}$ where, F - Force, K - spring constant, c - constant value, x - Stretch of the rope.
This rope isn't a simple $F=-kx$ spring for which we already have a potential energy formula. Because the rope force varies with stretch, we'll have to integrate. We can use the following equation for stored potential energy.
$\Delta U=-\int_{x1}^{x2}F(x)\space dx$ ; Let's plug known values into this equation.
$U=-\int_0^x(-kx-cx^{2})dx=\frac{1}{2}kx^{2}-\frac{c}{3}x^{3}\Biggr |_{0}^{x} $
$U=\frac{1}{2}kx^{2}-\frac{1}{3}cx^{3}-(1)$
Here given that, when the rope is stretched into the distance 'd' its potential energy is twice what it would be if the rope were an ideal spring with $F=-kx.$ So we can write,
$2\times\frac{1}{2}kx^{2}=\frac{1}{2}kx^{2}-\frac{1}{3}cx^{3} $
$\frac{c}{3}x=\frac{k}{2}$; Let's plug known values into this equation
$\frac{cd}{3}=\frac{k}{2}=\gt c=\frac{3k}{2d}$
