1. Parametrize the parabola

For the parabola $y^2=6x$, we compare with the standard form $y^2=4ax$. Thus, $4a=6 \implies a=\frac{3}{2}.$

A general point on the parabola is $P(t)=\left(at^2,2at\right)=\left(\frac{3}{2}t^2,3t\right).$

Let $A\equiv \left(\frac{3}{2}t_1^2,3t_1\right),\quad B\equiv \left(\frac{3}{2}t_2^2,3t_2\right),\quad C\equiv \left(\frac{3}{2}t_3^2,3t_3\right).$

2. Equation of line $AB$

For parabola $y^2=4ax$, the chord joining parameters $t_1,t_2$ is $y(t_1+t_2)=2x+2at_1t_2.$

Here $a=\frac{3}{2}$, so $y(t_1+t_2)=2x+3t_1t_2.$

3. Coordinates of point $D$

The line $L$ through $C$ parallel to the $x$-axis is $y=3t_3.$

Since $D$ is the intersection of $AB$ with $L$, substitute $y=3t_3$ into the chord equation: $3t_3(t_1+t_2)=2x_D+3t_1t_2.$ So, $x_D=\frac{3}{2}\Big(t_3(t_1+t_2)-t_1t_2\Big).$

Hence $D=\left(\frac{3}{2}(t_3(t_1+t_2)-t_1t_2),3t_3\right).$

4. Lengths $AM$ and $BN$

Since $L$ is horizontal, the perpendiculars from $A$ and $B$ to $L$ are vertical. So their lengths are just vertical distances from $A$ and $B$ to the line $y=3t_3$: $AM=|3t_1-3t_3|=3|t_1-t_3|,$ $BN=|3t_2-3t_3|=3|t_2-t_3|.$

Therefore, $AM\cdot BN=9|(t_1-t_3)(t_2-t_3)|.$

5. Length $CD$

Since both $C$ and $D$ lie on the horizontal line $L$, $CD$ is the horizontal distance: $CD=|x_D-x_C|.$

Now $x_C=\frac{3}{2}t_3^2.$ So, $CD=\left|\frac{3}{2}(t_3(t_1+t_2)-t_1t_2)-\frac{3}{2}t_3^2\right|.$ Factor: $CD=\frac{3}{2}\left|t_3t_1+t_3t_2-t_1t_2-t_3^2\right|.$ Now, $t_3t_1+t_3t_2-t_1t_2-t_3^2=(t_1-t_3)(t_3-t_2)=-(t_1-t_3)(t_2-t_3).$ Hence, $CD=\frac{3}{2}|(t_1-t_3)(t_2-t_3)|.$

6. Compute the required expression

$\frac{AM\cdot BN}{CD} = \frac{9|(t_1-t_3)(t_2-t_3)|}{\frac{3}{2}|(t_1-t_3)(t_2-t_3)|}=\frac{9}{3/2}=6.$

Therefore, $\left(\frac{AM\cdot BN}{CD}\right)^2=6^2=36.$

7. Final answer

The required integer is $\boxed{36}.$