Answer
$2.5\times10^3\;N$. No, it is not possible.
Work Step by Step
By symmetry, we know that the two tensions have the same magnitude. The rope sags at an angle $\theta$ below the horizontal, where $sin\;\theta = \frac{2.1\;m}{18m}$, so $\theta = 6.70^{\circ}$.
Write Newton’s second law for the vertical direction.
$$\Sigma F_y=2\times F_T sin\;6.70^{\circ}-mg=0$$
$$F_T=\frac{mg}{ 2\;sin\;6.70^{\circ}}=\frac{(60.0\;kg)(9.80\;m/s^2)}{ 2\;sin\;6.70^{\circ}}=2.5\times10^3\;N$$
The larger the tension gets, the less the wire sags. However, it is impossible to increase the tension enough to eliminate the sag entirely. There is always going to be a vertical component of the tension to balance the person’s (and the cable’s) weight.
