I'm guessing "distance from the plane to the station" refers to the Euclidean distance, or how far away the plane is in space from the station, and not horizontal distance, or how far it would be from the station if it were traveling at the same altitude as the station.
Let [tex]d[/tex] be the distance from the plane to the station. The plane forms a right triangle with the station with one leg of length 1 mi. The second leg's length [tex]x[/tex] can be determined from by the plane's current distance [tex]d[/tex] via the Pythagorean theorem:
[tex]1^2+x^2=d^2[/tex]
So when the plane's distance is [tex]d[/tex], we have [tex]x=\sqrt{4^2-1^2}=\sqrt{15}[/tex].
We know that [tex]x[/tex] changes at the same rate as the plane's speed, i.e. [tex]\dfrac{\mathrm dx}{\mathrm dt}=520\text{ mi/h}[/tex]. Differentiating the equation above with respect to time [tex]t[/tex], we have
[tex]2x\dfrac{\mathrm dx}{\mathrm dt}=2d\dfrac{\mathrm dd}{\mathrm dt}[/tex]
At the point when [tex]d=4[/tex], we found that [tex]x=\sqrt{15}[/tex], and we're given [tex]\dfrac{\mathrm dx}{\mathrm dt}=520[/tex], so we find
[tex]\dfrac{\mathrm dd}{\mathrm dt}=\dfrac{2\sqrt{15}\cdot520}{2\cdot4}=130\sqrt{15}\approx504\text{ mi/h}[/tex]
