I find 16,000 ft above with the approximate 1/60 rule : (+2°+2°) * (1/60°) * 40 nm * (6000 ft/nm) = +16,000 ft
Are you sure that this question is not a mix of 2 different questions/answers pairs?
I found the following very similar question on the internet:
#48. The tilt angle on the AWR at which an active cloud just disappears from the screen is 4 degrees up. If the beam width is 5 degrees and the range to the cloud is 40 nm use the 1 in 60 rule to calculate the approximate height of the cloud relative to the aircraft?
4000 ft above.
4000 ft below.
6000 ft above.
6000 ft below.
For that question, the answer is : (+4°- 2.5°) * (1/60°) * 40 nm * (6000 ft/nm) = + 6,000 ft , or 6000 ft above
Post Scriptum: bear in mind that "disappears" practically means "fades with a 50% return" as the beam width is defined as the -3 dB perimeter of the outgoing wave (-3dB = 50% signal).