Solution:

On the first day, the snail climbs up 7 ft and slips down 4 ft while sleeping. So, next morning, it is 3 ft from where it started. The snail thus travels 3 ft upwards every day. Therefore, in 3 days, it has traveled a distance of 9 ft from the bottom.

Here lies the catch to the problem! On the last day, the snail travels 7 ft upwards and hence reaches the top of the wall in a total of 4 days.

Alternative Solution through Equation:

Let x be the number of days the snail takes to reach the top of the wall 16 ft in height.

On the last day, the snail will reach the top by traveling 7 ft upwards and there will not be any question of slipping down.
The number of remaining days excluding the last day are (x − 1). Since the snail climbs up 7 ft and slips down 4 ft while sleeping, it travels 3 ft upwards on each of these remaining days. Thus,

Distance traveled on last day + Distance traveled on remaining days = Wall height; or
7 + 3 (x − 1) = 16

On solving the above equation, we get

3 (x − 1) = 16 − 7 = 9; or
x = (9 / 3) + 1 = 4.