Answer
The escape speed is 42.1 km/s.
Work Step by Step
To find the escape speed, we can assume that the speed is zero when the probe is very far away. Also, the gravitational potential energy is zero when the probe is very far away.
We can use conservation of energy to find the escape speed. Let $M_s$ be the sun's mass and let $M_p$ be the probe's mass.
$K_0+U_0 = K_f+U_f$
$\frac{1}{2}M_pv_0^2-\frac{G~M_s~M_p}{R_0} = 0+0$
$\frac{1}{2}M_pv_0^2=\frac{G~M_s~M_p}{R_0}$
$v_0^2=\frac{2~G~M_s}{R_0}$
$v_0=\sqrt{\frac{2~G~M_s}{R_0}}$
$v_0=\sqrt{\frac{(2)(6.67\times 10^{-11}~m^3/kg~s^2)(1.99\times 10^{30}~kg)}{1.50\times 10^{11}~m}}$
$v_0 = 4.21\times 10^4~m/s$
$v_0 = 42.1~km/s$
With an initial speed of 42.1 km/s, the probe will be able to escape from the solar system. The escape speed is 42.1 km/s.
