5. The oscillations of the ball consist of two stages: in water and in air. When the ball enters the water, a constant Archimedes' force begins to act on it, which does not change the frequency of the spring pendulum oscillations but shifts the equilibrium position.

Upon entering the water, the spring is stretched by \( \frac{mg}{k} \), and in the equilibrium position (taking into account the Archimedes' force), it is stretched by \( \frac{mg(1 - \rho_0 / \rho)}{k} \). Thus, the spring extension is \( \frac{mg}{k} \frac{\rho_0}{\rho} \).

The oscillations are described by the equation \( x = A \cos(\omega t - \varphi) \), with the velocity \( V = -\omega A \sin(\omega t - \varphi) \), and \( \omega = \sqrt{\frac{k}{m}} \). Substituting the initial conditions \( \left( x = \frac{mg}{k} \frac{\rho_0}{\rho}, V = v \right) \). Hence, \( \varphi = \operatorname{arctg} \frac{kv \rho}{\omega mg \rho_0} = \operatorname{arctg} \left( \frac{v \rho}{g \rho_0} \sqrt{\frac{k}{m}} \right) \).

The ball remains in the water for one period while \( A \cos(\omega t - \varphi) > \frac{mg}{k} \frac{\rho_0}{\rho} \), i.e., \( 0 < t < \frac{2 \varphi}{\omega} \). In the air, the ball spends half a period of the spring pendulum oscillations.
Answer: $\sqrt{\frac{m}{k}}\left[\pi+2 \operatorname{arctg}\left(\frac{v \rho}{g \rho_{0}} \sqrt{\frac{k}{m}}\right)\right]$

In all options, the answer is the same.