Let, $\mathrm{XY}$ - direction of the bat $\mathrm{AO}$-direction of the ball before the strike
$\mathrm{OB}$ - direction of the ball after the strike
$\mathrm{OD}-$ normal to the bat $\mathrm{XY}$
$$
\therefore \quad \alpha=\angle \mathrm{DOA}=45 / 2=22.5^{\circ}
$$
Let, $u$ - initial velocity of the ball along AO, $u \cos \alpha$ - component of the velocity along DO
$u \sin \alpha$-component of the velocity along $\mathrm{XY}$
From the figure, it is clear that velocity along horizontal direction is just changing its direction not the magnitude.
$\therefore$ Impulse on the ball $=m u \cos \alpha-(-m u \cos \alpha)$
$=2 m u \cos \alpha=2 \times 0.15 \times 15 \cos 22.5^{\circ}$
$=4.16 \mathrm{kgms}^{-1}$.