1. Impulse along a direction

Impulse equals change in momentum. Along the $x$-direction,

$J_x = \Delta p_x = m(v_{fx}-v_{ix})$

When a ball strikes a rigid wall and is reflected with the same speed, the component of velocity parallel to the wall remains unchanged, while the component perpendicular to the wall reverses.

So the magnitude of impulse is:

$|J_x| = 2m|v_x|$

provided $x$ is normal to the wall.

2. Convert speed

Given speed:

$108\ \text{km/h} = 30\ \text{m/s}$

Mass of each ball:

$m = 30\text{ g} = 0.03\text{ kg}$

Since mass and speed are same for both balls, the ratio of impulses depends only on the normal $x$-component of velocity.

3. For ball $a$

From the figure, ball $a$ strikes the wall at angle $45^\circ$ with the normal ($x$-direction). Hence

$v_{x,a} = v\cos 45^\circ = \frac{v}{\sqrt{2}}$

Therefore,

$|J_{x,a}| = 2m\frac{v}{\sqrt{2}}$

4. For ball $b$

From the figure, ball $b$ strikes the wall normally along the $x$-direction, so

$v_{x,b} = v$

Hence,

$|J_{x,b}| = 2mv$

5. Take the ratio

$\frac{|J_{x,a}|}{|J_{x,b}|} = \frac{2m\frac{v}{\sqrt{2}}}{2mv} = \frac{1}{\sqrt{2}}$

So,

$|J_{x,a}| : |J_{x,b}| = 1 : \sqrt{2}$

6. Final answer

The required ratio is

$\boxed{1 : \sqrt{2}}$

So the correct option is D.

7. Comparison with stored answer

Stored correct answer is B: $\sqrt{2}:1$.

My derived answer is D: $1: \sqrt{2}$.

Hence, I do not agree with the stored answer. The stored answer appears to have the ratio inverted.