The field is a rectangle of length 68.5 m and breadth 45.6 m. A path of uniform width 2.4 m runs all around the outside of the field, so the outer boundary of the path is itself a larger rectangle, concentric with the field.

Step 1 — Perimeter of the original field.
Perimeter = 2 × (length + breadth) = 2 × (68.5 + 45.6) = 2 × 114.1 = 228.2 m.

Step 2 — Outer dimensions of the path. Because the path is added on both sides of each dimension, each side grows by 2 × 2.4 = 4.8 m:

• Outer length = 68.5 + 4.8 = 73.3 m
• Outer breadth = 45.6 + 4.8 = 50.4 m

Step 3 — Outer perimeter.
= 2 × (73.3 + 50.4) = 2 × 123.7 = 247.4 m.

Step 4 — Difference.
247.4 − 228.2 = 19.2 m.

A useful shortcut: for any rectangle, adding a uniform border of width w all around increases the perimeter by exactly 8w, no matter what the original length and breadth are. Here 8 × 2.4 = 19.2 m, which matches instantly. This is because each of the four corners contributes an extra 2w to the boundary path. Notice the answer does not depend on the numbers 68.5 or 45.6 at all.

The values 17 m, 20 m and 18.5 m do not correspond to 8 × 2.4; they would arise only from arithmetic slips such as adding the width on one side only (which would give 4w) or mis-computing the increment. The correct difference is 19.2 m.