Find the area of inscribed convex octagon, if the length of four sides is $2$, and length of other four sides is $ 6\sqrt {2}$.

$\textbf{(A)}\ 120 \qquad\textbf{(B)}\  24 \plus{} 68\sqrt {2} \qquad\textbf{(C)}\  88\sqrt {2} \qquad\textbf{(D)}\ 124 \qquad\textbf{(E)}\  72\sqrt {3}$
1. **Identify the structure of the octagon:**
   The octagon is inscribed, meaning it is inside a circle. The sides are arranged in the pattern \(2, 6\sqrt{2}, 2, 6\sqrt{2}, 2, 6\sqrt{2}, 2, 6\sqrt{2}\).

2. **Break down the octagon into simpler shapes:**
   - The octagon can be divided into four \(45^\circ-45^\circ-90^\circ\) triangles and four rectangles.
   - Each \(45^\circ-45^\circ-90^\circ\) triangle has legs of length \(6\) (since the hypotenuse is \(6\sqrt{2}\)).
   - Each rectangle has dimensions \(6 \times 2\).

3. **Calculate the area of the triangles:**
   - The area of one \(45^\circ-45^\circ-90^\circ\) triangle is given by:
     \[
     \text{Area of one triangle} = \frac{1}{2} \times 6 \times 6 = 18
     \]
   - Since there are four such triangles:
     \[
     \text{Total area of triangles} = 4 \times 18 = 72
     \]

4. **Calculate the area of the rectangles:**
   - The area of one rectangle is:
     \[
     \text{Area of one rectangle} = 6 \times 2 = 12
     \]
   - Since there are four such rectangles:
     \[
     \text{Total area of rectangles} = 4 \times 12 = 48
     \]

5. **Calculate the area of the central square:**
   - The central square has side length \(2\):
     \[
     \text{Area of the square} = 2 \times 2 = 4
     \]

6. **Sum the areas of all the shapes:**
   \[
   \text{Total area} = \text{Total area of triangles} + \text{Total area of rectangles} + \text{Area of the square}
   \]
   \[
   \text{Total area} = 72 + 48 + 4 = 124
   \]

The final answer is \(\boxed{124}\).