Solution

Potential energy of the weight:

$$E_p = -mgx_0$$

Potential energy of rubber cord

$$E_p' = k' \frac{\left(\sqrt{l^2 + x_0^2} - l\right)^2}{2}$$

From the law of conservation of energy, the total energy of the system must equal zero

$$E = 0 = E_p + E_p' + E_p'$$

It follows from the figure that the stretching of the cord is

$$\Delta L = \sqrt{l^2 + x_0^2} - l$$

Since we now consider them as separate springs, cords, we use superposition:

$$\frac{1}{k} = \frac{1}{k'} + \frac{1}{k'} \implies k' = 2k$$

From where stiffness of the cord could be found as

$$mgx_0 = \frac{2 \cdot 2k \cdot \left(\sqrt{l^2 + x_0^2} - l \right)^2}{2}$$

$$\boxed{k = \frac{mgx_0}{2 \cdot \left( \sqrt{l^2 + x_0^2} - l \right)^2}}$$