Solution

Important thing - in the center of an asteroid the body's gravitational potential energy will not be zero. Each layer of the asteroid will affect the body. We need to find the gravitational potential in the center for each layer:
$$\phi_G=-\int\frac{\gamma dm}{r}$$
The mass of each layer we can find with common equation of density(due to the homogeneity of the asteroid)
$$dm=\rho dV$$
Density of ball:
$$\rho =\frac{3m}{4\pi R^3}$$
Derivative of the volume of a sphere with respect to its radius
$$dV=4\pi r^2dr$$
let's substitute this into the potential equation
$$\phi_G=-4\pi\gamma \rho\int rdr$$
integration is carried out in the range from zero to $R$
$$\phi_G=-4\pi\gamma \rho\frac{R^2}{2}$$
let's substitute the density
$$\phi_G=-\frac{3\gamma m}{2R}$$
in order for the body to leave the gravitational fields of the asteroid, the amount of mechanical energy must be equal to zero
$$M\phi+\frac{Mv^2}{2}=0$$
from here we can find the square of the velocity
$$v^2=\frac{3\gamma m}{R}$$
hence the speed itself
$$\boxed{v=\sqrt{\frac{3\gamma m}{R}}}$$