Given, Side of the square field $= 10\,m$

Therefore, perimeter $= 10 \,m \times 4 = 40 \,m$

Farmer moves along the boundary in $40\,s.$

Displacement after $2\, m$ $20\, s = 2 \times 60\, s + 20\, s = 140\, s =$ ?

Since in $40 \,s$ farmer moves $40 \,m$

Therefore, in $1\,s$ distance covered by farmer $= 40 / 40 \,m = 1\,m$

Therefore, in $140\,s$ distance covered by farmer $= 1 \times 140 \,m = 140 \,m$

Now, number of rotation to cover $140$ along the boundary

$=$ Total Distance $/$ Perimeter 

$= 140 \,m / 40\, m = 3.5$ round

Thus, after $3.5$ round farmer will at point $C$ of the field.

Therefore. Displacement $A C=\sqrt{(10 \,m)^{2}+(10 \,m)^{2}}$

$=\sqrt{100 \,m^{2}+100\, m^{2}}$

$=\sqrt{200 \,m^{2}}$

$=10 \sqrt{2} \,m$

$=10 \times 1.414=14.14 \,m$

Thus, after $2$ $min$ $20\,\sec$ the displacement of farmer will be equal to $14.14 \,m$ north east from intial position.