Answer
$J = (0.45~kg~m/s)~\hat{i}+(0.45~kg~m/s)~\hat{j}+(1.08~kg~m/s)~\hat{k}$
Work Step by Step
In part (b), we found that the impulse on the ball from the wall is:
$J = (-0.45~kg~m/s)~\hat{i}-(0.45~kg~m/s)~\hat{j}-(1.08~kg~m/s)~\hat{k}$
By Newton's Third Law, the impulse on the wall from the ball is equal in magnitude, but opposite in direction, to the impulse on the ball from the wall.
Therefore, the impulse on the wall is:
$J = (0.45~kg~m/s)~\hat{i}+(0.45~kg~m/s)~\hat{j}+(1.08~kg~m/s)~\hat{k}$
