Its question about velocity time graph?

Magnitude of Frictional force acting on the block `=0.12\ N`.
We know that force of friction always opposes motion. As such when the block moves from `X` to `Y`. The frictional force produces a retardation. Which can be found from Newton's Second Law of Motion.

`a=-F/m=(-0.12)/0.15=-0.8\ ms^-2`

To find the velocity `v` of block when it hits `Y` can be found from the kinematic expression

`v=u+at` ........(1)

Inserting given values we get

`v=3+(-0.8xx2)`
`v=1.4\ ms^-1` ......(2)

The block rebounds from the wall and moves directly towards `X`
before coming to rest at the point `Z`.

Kinetic energy as it his the wall `=1/2mv^2=1/2xx0.15xx(1.4)^2`
`=0.147\ J`
Kinetic energy lost on hitting the wall`=0.072\ J`.
Remaining kinetic energy on rebound`=0.147-0.072=0.075\ J`
Velocity `v_r` as it leaves `Y` on rebound can be found from

`1/2mv_r^2=0.075`
`=>v_r=sqrt(2xx0.075/0.15)=+-1\ ms^-1`
Selecting negative root as now the velocity is in opposite direction of initial velocity.
`v_r=-1\ ms^-1` .......(3)

Time when block comes to rest at `Z` on rebound can be found from kinematic expression (1).
We note that after rebound magnitude of frictional force remains same but its direction changes. Hence we have

`0=-1+0.8t`
`=>t=1/0.8=1.25\ s` ......(4)

Once we have acceleration, velocity, and time we can sketch the velocity-time graph as shown below.