Note: The question did not mention whether the climb gradient of 4.5% is in still air, or after the wind has been applied. They have given us a height difference formula, though, which includes a correction for TAS being different from groundspeed, and therefore a wind correction will apply itself automatically in that formula, so we can assume the 4.5% is in still air, which is also what you would usually get from using a climb gradient graph or calculation.

The height difference formula that they have given us is a lovely addition to this question, and makes it a lot more simple.

Calculating the head/tailwind requires us to use trigonometry, or one of the wind component charts like in the CAP. The headwind (in this case) is the windspeed* multiplied by cosine of the angle difference from runway heading, so as our wind is 230º and runway is 27 (270º), the difference is 40º, meaning that the headwind component is 14 x cos(40º) = 10.7 knots.

*Note that the full windspeed is used, as we are trying to make an actual height calculation here, not a legal minimum height as we would do in MEP and class A aircraft. SEP aircraft have no obstacle clearance requirements, so there are no such "50% of headwind, 150% of tailwind" rules to apply.

This means that the groundspeed is 115 - 10.7 = 104.3 kts.

The ground distance off the figure before the obstacle (control tower) is 3 NM, so 3 x 6080 = 18 240 ft

Using the formula then, the height difference is ((18240 x 4.5)/ 100) x (115/104.3) = 905 ft

This is the height difference, and is added to the 50 ft screen height which is assumed to be gained at the end of the runway, so the total altitude upon reaching the obstacle is 955 ft.

Minus the obstacle height of 200 ft, this means that the obstacle clearance is 755 ft, closest answer 753 ft.