Answer
The impulse on the ball from the wall is $~~(1.8~kg~m/s)~\hat{j}$
Work Step by Step
Since the initial angle is the same as the rebound angle while the speed stays the same, the x-component of the velocity (and momentum) does not change.
We can find the change in the y-component of the velocity:
$\Delta v_y = v_{yf} - v_{yi}$
$\Delta v_y = (6.0~m/s)~sin~30^{\circ}-[-(6.0~m/s)~sin~30^{\circ}]$
$\Delta v_y = (6.0~m/s)~\hat{j}$
We can find the change in momentum:
$\Delta p = m~\Delta v$
$\Delta p = (0.300~kg)~(6.0~m/s)~\hat{j}$
$\Delta p = (1.8~kg~m/s)~\hat{j}$
The impulse on the ball is equal to the ball's change in momentum.
Therefore, the impulse on the ball from the wall is $~~(1.8~kg~m/s)~\hat{j}$.
