Let the proper time interval measured by the father in his own frame be $\Delta t_0$. The total time for the journey in his frame is: $$ \Delta t_0 = 2 \text{ years (outward)} + 2 \text{ years (return)} = 4 \text{ years} $$ Let $\Delta t$ be the time interval measured by the daughter on Earth. According to the problem, the father finds himself 20 years younger than his daughter upon returning. Since the father was already 20 years older than his daughter, the total time difference accumulated by the daughter relative to the father is: $$ \Delta t = 4 \text{ years} + 20 \text{ years} + 20 \text{ years} = 44 \text{ years} $$ Using the time dilation formula, where $v$ is the velocity of the spaceship and $c$ is the speed of light: $$ \Delta t = \frac{\Delta t_0}{\sqrt{1 - \frac{v^2}{c^2}}} $$ Substituting the values: $$ \begin{aligned} 44 &= \frac{4}{\sqrt{1 - \frac{v^2}{c^2}}} \\ \sqrt{1 - \frac{v^2}{c^2}} &= \frac{4}{44} = \frac{1}{11} \\ 1 - \frac{v^2}{c^2} &= \left( \frac{1}{11} \right)^2 \\ \frac{v^2}{c^2} &= 1 - \frac{1}{121} \\ \frac{v^2}{c^2} &= \frac{120}{121} \\ \frac{v}{c} &= \sqrt{\frac{120}{121}} \approx 0.9958 \end{aligned} $$ Thus, the velocity of the spaceship is: $$ v \approx 0.9958c $$ Taking $c = 3 \times 10^8 \text{ m/s}$ (or $2.998 \times 10^8 \text{ m/s}$ as per standard convention): $$ v \approx 0.9958 \times 3 \times 10^8 \text{ m/s} \approx 2.987 imes 10^8 \text{ m/s} $$