The De-orbit Burn

Question

The Shuttle must reduce its velocity at a pre-calculated point in its orbit in order to return to Earth. In order to reduce the velocity and change the orbit of the Shuttle, a maneuver called the de-orbit burn is performed. For this maneuver, the Shuttle is turned in a direction such that the Orbital Maneuvering System (OMS) nozzles point in the direction of the Shuttle's velocity back toward Earth. The OMS engines fire and give the Shuttle a velocity in the opposite direction, thus slowing the spacecraft.

The Shuttle must perform the de-orbit burn to change its orbit so that the perigee, the point in the orbit closest to Earth, is inside of Earth's atmosphere. De-orbit maneuvers are done to lower the perigee of the orbit to 60 miles (or less). An altitude of 60 miles is important because this is where the orbiting spacecraft is recaptured by Earth’s gravity and re-enters Earth’s atmosphere.

Calculate the minimum change in velocity (delta V or ∆V) required for the Space Shuttle to decrease its altitude to 60 miles if it’s orbiting with an apogee of 260 miles and a perigee of 220 miles above the surface of Earth.

Use the rule of thumb that below an altitude of 500 miles, for every 2 feet per second (ft/s) change in the orbiting space craft’s velocity its altitude will change by 1 mile.

and I have to answer in feet per second

Here's what I have-

change of altitude: 220-60=160
rule of thumb: 160mi.*2ft/sec=320ft/sec.

I don't know what to do from there