The given situation can be represented as shown in the following figure.

Where,

AO = Incident path of the ball

OB = Path followed by the ball after a deflection

∠AOB = Angle between the incident and deflected paths of the ball = 45°

∠AOP = ∠BOP = 22.5° = θ

Initial and final velocities of the ball = v

Horizontal component of the initial velocity = vcos θ along RO

Vertical component of the initial velocity = vsin θ along PO

Horizontal component of the final velocity = vcos θ along OS

Vertical component of the final velocity = vsin θ along OP

The horizontal components of velocities suffer no change. The vertical components of velocities are in the opposite directions.

∴Impulse imparted to the ball = Change in the linear momentum of the ball

$= \mathrm{mv} \cos \mathrm{\theta } -\left(-\mathrm{mv} \cos \mathrm{\theta }\right)$
$= 2 \mathrm{mv} \cos \mathrm{\theta }$

Mass of the ball, m = 0.15 kg

Velocity of the ball, v = 54 km/h = 15 m/s

Thus,

Impulse = 2 × 0.15 × 15 cos 22.5° = 4.16 kg m/s