A surveyor has to measure the horizontal distance from her position to a distant reference point C. Using her position as the center, a 200 m horizontal line segment is drawn with the two endpoints A and B. Points A, B and C are not collinear. Each of the angles ∠CAB and ∠CBA are measured as 87.8°. The distance (in m) of the reference point C from her position is nearest to

The correct option is 2603.

Let the position of the surveyor be $O$. Since the horizontal line segment $AB$ of length 200 m is drawn with the surveyor's position $O$ as the center, $O$ is the midpoint of $AB$.
Therefore, the lengths of the segments $OA$ and $OB$ are:

$$OA=OB=\frac{200}{2}=100\text{m}$$

We are given that the angles $\angle CAB=\angle CBA=87.8°$. Since the base angles are equal, triangle $ABC$ is an isosceles triangle with $AC=BC$.
In an isosceles triangle, the line joining the vertex $C$ to the midpoint $O$ of the base $AB$ is perpendicular to the base. Thus, $CO\perp AB$, and triangle $AOC$ is a right-angled triangle at $O$.

In the right-angled triangle $AOC$, we can apply trigonometric ratios to find the distance $OC$ of the reference point $C$ from the surveyor's position $O$:

$$\tan (\angle CAB)=\frac{OC}{OA}$$

Substituting the known values:

$$\tan (87.8°)=\frac{OC}{100}$$

Solving for $OC$:

$$OC=100\times \tan (87.8°)$$

Using the trigonometric value $\tan (87.8°)\approx 26.0307$:

$$OC\approx 100\times 26.0307=2603.07\text{m}$$

Rounding to the nearest integer, the distance of the reference point $C$ from her position is nearest to 2603 m.