The length of the chord of the parabola $$x^2 = 4y$$ having equation $$x - \sqrt{2}y + 4\sqrt{2} = 0$$ is:

We are given the parabola $$x^{2}=4y$$ and the line $$x-\sqrt{2}\,y+4\sqrt{2}=0$$. The points where this line meets the parabola will form the required chord; so we first find their coordinates by solving the two equations simultaneously.

From the line we can express $$y$$ in terms of $$x$$. Rearranging we get

$$x-\sqrt{2}\,y+4\sqrt{2}=0 \quad\Longrightarrow\quad \sqrt{2}\,y=x+4\sqrt{2}\quad\Longrightarrow\quad y=\dfrac{x+4\sqrt{2}}{\sqrt{2}}\;.$$

Now we substitute this value of $$y$$ in the parabola $$x^{2}=4y$$. Thus we have

$$x^{2}=4\left(\dfrac{x+4\sqrt{2}}{\sqrt{2}}\right).$$

Next we simplify the right-hand side. Since $$4/\sqrt{2}=\dfrac{4\sqrt{2}}{2}=2\sqrt{2}$$, the equation becomes

$$x^{2}=2\sqrt{2}\,x+16.$$

Bringing every term to the left, we obtain the quadratic equation

$$x^{2}-2\sqrt{2}\,x-16=0.$$

To solve it we use the quadratic‐formula $$x=\dfrac{-b\pm\sqrt{b^{2}-4ac}}{2a}$$, where here $$a=1,\,b=-2\sqrt{2}$$ and $$c=-16$$. Hence

$$x=\dfrac{2\sqrt{2}\pm\sqrt{(2\sqrt{2})^{2}-4(1)(-16)}}{2}.$$

Calculating inside the square root,

$$(2\sqrt{2})^{2}=4\cdot2=8,\qquad -4ac=-4(1)(-16)=64,$$

so

$$b^{2}-4ac=8+64=72\quad\Longrightarrow\quad\sqrt{72}=6\sqrt{2}.$$

Thus

$$x=\dfrac{2\sqrt{2}\pm6\sqrt{2}}{2}.$$

Separating the two possibilities,

$$x_{1}=\dfrac{2\sqrt{2}+6\sqrt{2}}{2}=\dfrac{8\sqrt{2}}{2}=4\sqrt{2},$$

$$x_{2}=\dfrac{2\sqrt{2}-6\sqrt{2}}{2}=\dfrac{-4\sqrt{2}}{2}=-2\sqrt{2}.$$

Corresponding $$y$$‐coordinates are obtained from $$y=\dfrac{x+4\sqrt{2}}{\sqrt{2}}$$.

For $$x_{1}=4\sqrt{2}$$ we get

$$y_{1}=\dfrac{4\sqrt{2}+4\sqrt{2}}{\sqrt{2}}=\dfrac{8\sqrt{2}}{\sqrt{2}}=8.$$

For $$x_{2}=-2\sqrt{2}$$ we get

$$y_{2}=\dfrac{-2\sqrt{2}+4\sqrt{2}}{\sqrt{2}}=\dfrac{2\sqrt{2}}{\sqrt{2}}=2.$$

Therefore the end‐points of the chord are $$P_{1}(4\sqrt{2},\,8)$$ and $$P_{2}(-2\sqrt{2},\,2).$$

Now we find the distance between these points. The distance formula is

$$\text{Distance}=\sqrt{(x_{1}-x_{2})^{2}+(y_{1}-y_{2})^{2}}.$$

We compute the differences:

$$x_{1}-x_{2}=4\sqrt{2}-(-2\sqrt{2})=6\sqrt{2},$$

$$y_{1}-y_{2}=8-2=6.$$

Substituting,

$$\text{Distance}=\sqrt{(6\sqrt{2})^{2}+6^{2}}=\sqrt{36\cdot2+36}=\sqrt{72+36}=\sqrt{108}.$$

Since $$108=36\cdot3$$, we have $$\sqrt{108}=6\sqrt{3}.$$

Hence the length of the chord is $$6\sqrt{3}$$ units.

Hence, the correct answer is Option A.