Answer
$v = 8.9\times 10^4~m/s$
Work Step by Step
We can find an expression for the gravitational potential energy at the center of mass:
$U = 2\times (-\frac{GMm}{R}) = -\frac{2GMm}{R}$
In order to escape, the total energy at the center of mass must be at least zero. We can find the minimum required speed:
$K+U = 0$
$K = -U$
$\frac{1}{2}mv^2= \frac{2GMm}{R}$
$v^2= \frac{4GM}{R}$
$v= \sqrt{\frac{4GM}{R}}$
$v= \sqrt{\frac{(4)(6.67\times 10^{-11}~N~m^2/kg^2)(3.0\times 10^{30}~kg)}{1.0\times 10^{11}~m}}$
$v = 8.9\times 10^4~m/s$
