This question checks how well you know the layout elements of a simple circular curve used in road and railway surveying. Instead of quoting the apex distance formula directly, we build it from the tangent length and a right triangle.
When two straights meet at the point of intersection (PI), the curve touches each straight at a tangent point. The tangent length $T$ from the PI to a tangent point is given by $T = R\tan(\Delta/2)$, where $R$ is the radius and $\Delta$ is the deflection angle.
Here $R = 1000$ m and $\Delta = 60^{\circ}$, so $\Delta/2 = 30^{\circ}$.
\[ T = 1000 \times \tan 30^{\circ} = 1000 \times 0.577350 = 577.350 \text{ m} \]Now look at the right triangle formed by the centre of the curve $O$, a tangent point, and the PI. The angle at $O$ between the radius to the tangent point and the line $OI$ (the bisector) is $\Delta/2 = 30^{\circ}$. The side opposite this angle is the tangent length $T$, and the side adjacent is the radius $R$, so this triangle gives the same relation $T = R\tan(\Delta/2)$, confirming the setup.
The hypotenuse of this triangle is the line $OI$, joining the centre to the point of intersection. Using the cosine of the same angle,
\[ OI = \frac{R}{\cos(\Delta/2)} = \frac{1000}{\cos 30^{\circ}} = \frac{1000}{0.866025} = 1154.701 \text{ m} \]The apex distance $E$ is the extra length beyond the curve along this bisector, measured from the PI back to the curve itself, so it is simply $OI$ minus the radius $R$.
\[ E = 1154.701 - 1000 = 154.701 \text{ m} \]Let's summarize:
So the apex distance of the circular curve is $154.701$ m.