Step-by-step Solution

1. Principle of Energy Conservation

   To escape the Sun-Earth system, the rocket must have enough initial kinetic energy to overcome the combined gravitational potential energy of the Earth and the Sun. The minimum velocity corresponds to the rocket reaching an infinite distance with zero kinetic energy. According to the principle of conservation of mechanical energy:

   $E_{initial} = E_{final}$

   For the minimum escape velocity, the total energy at infinity ($E_{final}$) is zero. Therefore, the initial total energy must also be zero.

   $K_i + U_i = 0$

   where $K_i$ is the initial kinetic energy and $U_i$ is the initial potential energy of the rocket.

2. Initial Kinetic and Potential Energy

   The initial kinetic energy of the rocket (mass $m$) launched with velocity $v_s$ from the Earth's surface is:

   $K_i = \frac{1}{2}mv_s^2$

   The initial potential energy is the sum of the potential energy due to the Earth's gravity and the Sun's gravity. The rocket is on the Earth's surface at a distance $R_E$ from its center. The problem states the rocket is launched "away from the sun, along the line joining the Sun and the Earth", so its distance from the Sun's center is $d + R_E$, where $d$ is the distance between the Earth's and Sun's centers. However, since $d \gg R_E$, we can approximate this distance as $d$.

   The potential energy due to Earth is: $U_E = -\frac{GM_Em}{R_E}$ The potential energy due to the Sun is: $U_S = -\frac{GM_Sm}{d}$

   The total initial potential energy is: $U_i = U_E + U_S = -\frac{GM_Em}{R_E} - \frac{GM_Sm}{d}$

3. Setting up the Escape Velocity Equation

   Substituting the expressions for $K_i$ and $U_i$ into the energy conservation equation:

   $\frac{1}{2}mv_s^2 - \frac{GM_Em}{R_E} - \frac{GM_Sm}{d} = 0$

   Canceling the rocket's mass $m$ and rearranging for $v_s^2$:

   $v_s^2 = \frac{2GM_E}{R_E} + \frac{2GM_S}{d}$

4. Relating to Earth's Escape Velocity

   The escape velocity from Earth's gravitational field alone, given as $V_c = 11.2 \, \text{km/s}$, is defined by:

   $V_c^2 = \frac{2GM_E}{R_E}$

   Substituting this into our equation for $v_s^2$:

   $v_s^2 = V_c^2 + \frac{2GM_S}{d}$

5. Using the Given Ratios

   The problem provides the following data:

   • Mass of the Sun, $M_S = 3 \times 10^5 M_E$
   • Distance to the Sun, $d = 2.5 \times 10^4 R_E$

   Now, we express the second term in terms of $V_c^2$:

   $\frac{2GM_S}{d} = \frac{2G(3 \times 10^5 M_E)}{2.5 \times 10^4 R_E}$

   We can rearrange this to isolate the expression for $V_c^2$:

   $\frac{2GM_S}{d} = \left( \frac{3 \times 10^5}{2.5 \times 10^4} \right) \left( \frac{2GM_E}{R_E} \right)$

   Calculating the numerical factor:

   $\frac{3 \times 10^5}{2.5 \times 10^4} = \frac{30}{2.5} = 12$

   So, the second term is:

   $\frac{2GM_S}{d} = 12 \, V_c^2$

6. Calculating the Final Velocity

   Substitute this result back into the equation for $v_s^2$:

   $v_s^2 = V_c^2 + 12 V_c^2 = 13 V_c^2$

   Now, take the square root to find $v_s$:

   $v_s = \sqrt{13} \, V_c$

   Using the given value $V_c = 11.2 \, \text{km/s}$:

   $v_s = \sqrt{13} \times 11.2 \, \text{km/s}$ $v_s \approx 3.606 \times 11.2 \, \text{km/s} \approx 40.38 \, \text{km/s}$

7. Comparing with Options

   The calculated minimum initial velocity is approximately $40.38 \, \text{km/s}$. We must find the closest value among the options:

   • A: $22 \, \text{km/s}$
   • B: $42 \, \text{km/s}$
   • C: $62 \, \text{km/s}$
   • D: $72 \, \text{km/s}$

   The value $40.38 \, \text{km/s}$ is closest to $42 \, \text{km/s}$. The small difference might be due to the rounding of constants given in the problem.