Scale = CD/ED = 1/9,260,000. If CD = 6 cm then ED = 6 x 9,260,000cm = 555.6km.
555.6km = 300nm and this is the departure distance between the two meridians 10º apart.

Departure = Ch.long(') x cos lat so cos lat = departure/ch.long(') = 300/600 = 0.5.
0.5 is cos 60º so this is at 60º N or S.