(N/A) An instinctive answer to $(i)$ might be that the force on the wall in case $(a)$ is normal to the wall,while that in case $(b)$ is inclined at $30^{\circ}$ to the normal. This answer is wrong. The force on the wall is normal to the wall in both cases.
To find the force on the wall,we consider the impulse on the ball due to the wall using the second law,and then use the third law to answer $(i)$. Let $u$ be the speed of each ball before and after collision with the wall,and $m$ be the mass of each ball. Choose the $x$ and $y$ axes as shown in the figure.
Case $(a)$:
Initial momentum: $(p_x)_{\text{initial}} = mu, (p_y)_{\text{initial}} = 0$
Final momentum: $(p_x)_{\text{final}} = -mu, (p_y)_{\text{final}} = 0$
Impulse is the change in momentum vector. Therefore,
$x$-component of impulse $= -2mu$
$y$-component of impulse $= 0$
Impulse and force are in the same direction. The force on the ball due to the wall is normal to the wall,along the negative $x$-direction. By Newton's third law,the force on the wall due to the ball is normal to the wall along the positive $x$-direction.
Case $(b)$:
Initial momentum: $(p_x)_{\text{initial}} = mu \cos 30^{\circ}, (p_y)_{\text{initial}} = -mu \sin 30^{\circ}$
Final momentum: $(p_x)_{\text{final}} = -mu \cos 30^{\circ}, (p_y)_{\text{final}} = -mu \sin 30^{\circ}$
Note that while $p_x$ changes sign,$p_y$ does not. Therefore,
$x$-component of impulse $= -2mu \cos 30^{\circ}$
$y$-component of impulse $= 0$
The direction of impulse (and force) is normal to the wall along the negative $x$-direction. By Newton's third law,the force on the wall is normal to the wall along the positive $x$-direction.
The ratio of the magnitudes of the impulses imparted to the balls in $(a)$ and $(b)$ is:
$\frac{2mu}{2mu \cos 30^{\circ}} = \frac{1}{\cos 30^{\circ}} = \frac{1}{\sqrt{3}/2} = \frac{2}{\sqrt{3}} \approx 1.15$.