Answer
$$v=177\text{ m}/\text{s}$$
Work Step by Step
The escape speed is the minimum speed an object needs to be able to "escape" from the gravity and It equals
$$v=\sqrt{\frac{2GM}{R}},$$
where $M$ is the mass of the body from the gravity of which we are escaping and $R$ is its radius.
Let's rewrite $M$ In terms of the density $\rho$ and volume $V$. For a sphere $V=\dfrac{4}{3}\pi R^3$.
$$M=\rho V=\rho \dfrac{4}{3}\pi R^3.$$
So $$\dfrac{M}{R}=\rho\dfrac{4}{3}\pi R^2.$$
The diameter of the asteroid is 300 km so $R=150$ km or $R=15\cdot 10^4\text{ m}$ and $\rho =2500$ $\text{kg}/\text{m}^3$, $G=6.67\cdot10^{-11}\text{N}\cdot \text{m}^2/\text{kg}^2$.
$$v=\sqrt{\dfrac{8\cdot 3.141}{3}\cdot 6.67\cdot10^{-11}\cdot 2500\cdot (15\cdot 10^4)^2}=177\text{ m}/\text{s}$$
