Only One Metre

Since people are misinterpreting the question, I will write a solution to show where they are going wrong.

Note that the string is tangent to the Earth from where it is being held to the point of first contact with the surface. These are the two tangents from the point to the Earth as a result. Let's call the lengths of these tangents $a$ . Let's call the height of the point from the ground $h$ and the radius of the Earth $r$ (we will substitute $r=6400000 \mathrm{m}$ later).

Consider a line joining the point to the centre of the Earth. This has a length of $r+h$ , which is equal to $\sqrt{a^2+r^2}$ by Pythagoras Theorem. Call the angle subtended by the line connecting the centre of the Earth to the point of tangency of the string and the line connecting the point to the centre of the earth as $\theta$ (in radians).

The length of the string not in contact with the Earth is $2a$ . Also, the length of the corresponding circular arc is $2r\theta$ . Thus, we have

$\begin{aligned} 2a &= 2r\theta + 1\\ a &= r\theta + \dfrac {1}{2}\\ \dfrac {a}{r} &= \theta + \dfrac {1}{2r}\\ \tan \theta &= \theta + \dfrac {1}{2r} \end{aligned}$

The Maclaurin series for $\tan \theta$ is $\theta+\dfrac{\theta^3}{3} + \ldots$ . Hence,

$\begin{aligned} \theta+\dfrac {\theta^3}{3} &\approx \theta + \dfrac {1}{2r}\\ \theta^3 &\approx \dfrac {3}{12800000}\\ \theta &\approx \dfrac {\sqrt[3]{15}}{400} \end{aligned}$

From this, we get the value of $a$ is

$\begin{aligned} a &= 6400000 \times \dfrac {\sqrt[3]{15}}{400} + \dfrac {1}{2}\\ &= 16000 \sqrt[3]{15} + \dfrac {1}{2} \end{aligned}$

From this we get the value of $r+h$ as

$\begin{aligned} r+h &= \sqrt{\left (16000 \sqrt[3]{15} + \dfrac {1}{2}\right )^2 + 64000000^2}\\ &\approx 6400121.6 \mathrm{m}\\ h &\approx 121.6 \mathrm{m} \end{aligned}$

Thus, the height is $121.6 \mathrm{m}$ .