What's the length of the string?

No need calculus to solve this problem; just geometry.

Imagine that you apply ink on the string and then roll it on a piece of paper. Then the string will trace a (blue) straight line print on the paper. Since the string has traversed a horizontal distance of $n \times n = n^2$ and a vertical distance of $3n$ . The hypotenuse of the right triangle is the length of the string $20$ . Therefore $\sqrt{n^4 + 9n^2} = 20 \implies n \sqrt{n^2+9} = 20 \implies n = \boxed 4$ .

Let $r$ be the radius of the rod.

$2\pi r = n \implies r = \dfrac{n}{2\pi}$

$\implies x = \dfrac{n}{2\pi}\cos(\theta), y = \dfrac{n}{2\pi}\sin(\theta)$ and $z = \dfrac{3}{2\pi}\theta$ , where $\theta \in [0,2\pi n]$ .

$\implies \dfrac{dx}{d\theta} = -\dfrac{n}{2\pi}\sin(\theta), \dfrac{dy}{d\theta} = \dfrac{n}{2\pi}\cos(\theta)$ and $\dfrac{dz}{d\theta} = \dfrac{3}{2\pi}$

$\implies$ the length of the string $L = \int_{0}^{2\pi n} \sqrt{(\dfrac{dx}{d\theta})^2 + (\dfrac{dy}{d\theta})^2 + (\dfrac{dz}{d\theta})^2} d\theta$

$= \int_{0}^{2\pi n} \dfrac{\sqrt{n^2 + 9}}{2\pi} d\theta = \dfrac{\sqrt{n^2 + 9}}{2\pi}\theta|_{0}^{2\pi n} = n\sqrt{n^2 + 9} = 20 \implies$

$n^2(n^2 + 9) = 400 \implies n^4 + 9n^2 - 400 = 0 \implies n^2 = 16,-25 \implies n = \pm 4, \pm 5i$

$\therefore n = \boxed{4}$ .