The area enclosed between the velocity-time sketch and the time axis gives:
A) Acceleration of a body
B) Displacement of a body
C) Distance travelled by a body
D) Nothing specifically

Hint: Velocity-time graphs are used to describe the motion of objects that are moving in a straight line. They can be used to show acceleration and perform displacement work.In the velocity-time graph, any area enclosed will be the product of velocity and time.

Complete solution:
As we know that the unit of
${\text{velocity,V = }}\dfrac{{\text{m}}}{{{\text{sec}}}}$
${\text{time, t = sec}}$
$\Rightarrow$ SI unit of $v \times t$
$\Rightarrow \dfrac{{{\text{metre}}}}{{{\text{sec}}}} \times \sec$
$\therefore {\text{ SI unit is metre}}$
Thus, velocity time graph gives the displacement
Hence ,the area enclosed in the velocity time graph is the displacement.
Which is the Area enclosed under the velocity-time graph .

So option B is correct.

Additional information:
This is the displacement and we consider the areas for negative velocities as negative.
The net area(considering both positive and negative areas) between the line and the time axis of the velocity-time graph is equal to the magnitude of displacement of the object.
When all areas are taken as positive then the total area under the velocity-time graph gives the distance covered by the body in a 1-d motion. Also that we are considering the motion to be along a straight line.

Note: Area under the velocity-time graph is displacement. And of course it is a vector quantity but area just gives the magnitude of displacement.
Direction can be found if direction of velocity vector is given, It is in the same direction to velocity if area is above x-axis.It is in opposite to velocity if area is below x-axis.