1. Resolve the initial velocity

Since the body is projected at $45^\circ$ with speed $u$,

$u_x = u\cos45^\circ = \frac{u}{\sqrt{2}}, \qquad u_y = u\sin45^\circ = \frac{u}{\sqrt{2}}$

2. Coordinates of the highest point

At the highest point, vertical velocity becomes zero.

Time to reach highest point:

$t_H = \frac{u_y}{g} = \frac{u/\sqrt{2}}{g} = \frac{u}{\sqrt{2}g}$

Horizontal distance of the highest point from the point of projection:

$x_H = u_x t_H = \frac{u}{\sqrt{2}}\cdot \frac{u}{\sqrt{2}g} = \frac{u^2}{2g}$

Vertical height of the highest point:

$y_H = \frac{u_y^2}{2g} = \frac{\left(\frac{u}{\sqrt{2}}\right)^2}{2g} = \frac{u^2/2}{2g} = \frac{u^2}{4g}$

So the position vector at the highest point is

$\vec r = x_H\hat i + y_H\hat j = \frac{u^2}{2g}\hat i + \frac{u^2}{4g}\hat j$

3. Velocity at the highest point

At the highest point, only horizontal velocity remains:

$\vec v = \frac{u}{\sqrt{2}}\hat i$

Hence momentum is

$\vec p = m\vec v = \frac{mu}{\sqrt{2}}\hat i$

4. Angular momentum about the point of projection

Angular momentum is

$\vec L = \vec r \times \vec p$

Using

$\vec r = x\hat i + y\hat j, \qquad \vec p = p_x\hat i$

we get

$\vec L = (x\hat i + y\hat j) \times (p_x\hat i)$

Since $\hat i \times \hat i = 0$,

$\vec L = y p_x (\hat j \times \hat i) = -y p_x \hat k$

So magnitude is

$L = y p_x$

Substitute values:

$L = \frac{u^2}{4g} \cdot \frac{mu}{\sqrt{2}} = \frac{mu^3}{4\sqrt{2}g}$

Now rewrite this in the given form:

$\frac{mu^3}{4\sqrt{2}g} = \frac{\sqrt{2}mu^3}{8g}$

Thus,

$\frac{\sqrt{2}mu^3}{Xg} = \frac{\sqrt{2}mu^3}{8g}$

Hence,

$X = 8$

5. Comparison with stored answer

Derived answer: $8$

Stored correct answer: $8$

They agree.