First step: Calculate the ISA temperature for each of the heights given.

20 000 ft	15 – 2 x 20 = -25ºC
18 000 ft	15 – 2 x 18 = -21ºC
16 000 ft	15 – 2 x 16 = -17ºC
14 000 ft	15 – 2 x 14 = -13ºC
12 000 ft	15 – 2 x 12 = -9ºC
10 000 ft	15 – 2 x 10 = -5ºC
8 000 ft	15 – 2 x 8 = -1ºC
6 000 ft	15 – 2 x 6 = 3ºC
4 000 ft	15 – 2 x 4 = 7º C
2 000 ft	15 – 2 x 2 = 11ºC
Surface	15ºC

Second step: We will now compare the above table (ISA temperature) to the actual temperature at those levels. In order to find out if we are dealing with conditions warmer, colder or equal to ISA.

• Looking at the question text, we conclude that the actual temperature is warmer than ISA. And we know that, disregarding pressure, for temperatures WARMER THAN ISA; TRUE ALTITUDE > INDICATED ALTITUDE

“The layer between 16 000 and 18 000 ft is absolutely unstable.” Incorrect – If ELR > DALR (3°C/1000 f) we have absolute instability. In this case, we have ELR < DALR.
“The temperature at 10 000 ft is in agreement with the temperature in the International Standard Atmosphere.” Incorrect – The actual temperature at 10 000 ft is -2ºC; ISA temperature at that same level is -5ºC. ISA +3ºC
“The height of the freezing level over the station is approximately 12 000ft.” incorrect – The freezing level is approximately at 9 000 ft. (8 000 ft = 2ºC, therefore 0ºC will be at 9 000 ft)