Find the area of the shaded region in the given figure.

(N/A) The outer rectangle $ABCD$ has length $L = 26 \, m$ and breadth $B = 12 \, m$.
Area of outer rectangle $= L \times B = 26 \times 12 = 312 \, m^2$.
The inner unshaded region consists of a rectangle and two semi-circles.
The breadth of the inner rectangle is $12 - (4 + 4) = 4 \, m$. This is also the diameter of the two semi-circles.
Radius of each semi-circle $r = 4 / 2 = 2 \, m$.
The length of the inner rectangle is $26 - (3 + 3 + 2 + 2) = 16 \, m$ (subtracting the $3 \, m$ gaps and the two radii of the semi-circles).
Area of inner rectangle $= 16 \times 4 = 64 \, m^2$.
Area of two semi-circles $= 2 \times (\frac{1}{2} \pi r^2) = \pi r^2 = \pi (2)^2 = 4\pi \, m^2$.
Total unshaded area $= 64 + 4\pi \, m^2$.
Area of shaded region $=$ Area of outer rectangle $-$ Total unshaded area
$= 312 - (64 + 4\pi) = (248 - 4\pi) \, m^2$.