If the space station is 7 miles in diameter, it is 3.5 miles in radius.
If we assume that we want to feel Earth-level artificial gravity at this distance from the center, we want the centripedal acceleration of a point on this cylinder to be the same as the acceleration due to gravity. In short,
[itex]a_{edge} = g[/itex]
but
[itex]a_{edge} = R \omega^{2}[/itex]
where R is the radius of the station (3.5 miles or 5607 meters) and omega is the angular velocity of the space station in radians per second.
Then we solve for [itex]\omega[/itex], finding that
[itex]\omega=\sqrt{\frac{g}{R}}[/itex]
so that [itex]\omega[/itex] is about 6.64thousandths of a revolution per second or about 0.40 revolutions per minute.