Alright, so

@wonderingwhy provided the formula for escape velocity and from there it’s just a matter of plugging a couple of values. G is the gravitational constant and it’s equal to about 6.67 * 10^-11. M is the mass of the star, which according to your link is not known exactly but estimated to be about 30 to 40 solar masses. A solar mass is about 2 * 10^30 kg, so if this star is, say, 35 solar masses, that’d put it at 70 * 10^30 kg. Finally, r is radius of the star, which according to the wikipedia article is about 1.5 * 10^9 km. We need that in meters to do this calculation, so add three more zeroes: 1.5 * 10^12 m.

So v = sqrt(2GM/r). Plug in the above values:

v = sqrt(2 * 6.67 * 10^-11 * 70 * 10^30 / 1.5 * 10^12)

v = 78,900 meters per second, or 176,500 miles per hour, or about 7 times more than earth’s escape velocity.

So why is it less than the sun’s? Because you chose the largest star by volume, not by mass. This star is only about 35 times more massive than the sun, but its radius is about 1000 times larger. Since escape velocity increases with the square root of mass/radius, having such a large radius means a smaller escape velocity even though the mass is also larger.

Interestingly, this must mean that this star has very low density. If it were the same density as the sun and had a radius 1000 times larger, its mass would be one billion times more than the sun’s (mass increases proportionate to the cube of radius). However, if it were this dense, I’m pretty sure it would be a black hole.