Answer:
Approximately [tex]4.16 \times 10^{3}\; {\rm m}[/tex] (or equivalently, [tex]4.16\; {\rm km}[/tex]) assuming this planet is spherical.
Explanation:
Let [tex]m[/tex] denote the mass of the ball. If the speed of the ball is [tex]v[/tex], the kinetic energy of the ball will be [tex](\text{KE}) = (1/2)\, m\, v^{2}[/tex].
Assume that the planet is spherical. Let [tex]r[/tex] denote the radius of the planet. Let [tex]M[/tex] denote the mass of this planet, and let [tex]G[/tex] denote the gravitational constant.
On the surface of this planet, the gravitational potential energy [tex](\text{GPE})[/tex] of this ball will be [tex](\text{GPE}) = (-G\, M\, m) / (r)[/tex]. (Note the negative sign. The ball is trapped inside the gravitational field of the planet, and it takes energy input to bring the ball out of this field.)
The ball is at its escape speed [tex]v_{e}[/tex] if the sum of [tex](\text{KE})[/tex] and [tex](\text{GPE})[/tex] at the surface of the planet is [tex]0[/tex]. In other words:
[tex](\text{KE}) + (\text{GPE}) = 0[/tex].
[tex]\begin{aligned}\frac{1}{2}\, m\, v^{2} + \frac{- G\, M\, m}{r} = 0\end{aligned}[/tex].
Rewrite this equation and solve for radius [tex]r[/tex].
[tex]\begin{aligned}\frac{G\, M\, m}{r} = \frac{1}{2}\, m\, v^{2}\end{aligned}[/tex].
[tex]\begin{aligned}r &= \frac{v^{2}}{2\, G\, M}\end{aligned}[/tex].
[tex]\begin{aligned}r &= \frac{2\, G\, M}{{v_{e}}^{2}} \\ &= \frac{2\, (6.67 \times 10^{-11}\; {\rm m^{3}\cdot kg^{-1} \cdot s^{-2})\, (1.56 \times 10^{15}\; {\rm kg})}}{(5.00\; {\rm m\cdot s^{-1}})^{2}} \\ &\approx 8.32 \times 10^{3}\; {\rm m}\end{aligned}[/tex].