Let's analyze the position function step by step.

The particle's position is given by:

$$x(t) = A \sin t + B \cos^2 t + C t^2 + D,$$

where $ t $ represents time, which has the dimension $ T $.

Since the overall dimensions of position $ x(t) $ must be length $ (L) $, each term in the expression must also have dimensions of length.

Term $ A \sin t $:

The sine function, $ \sin t $, is dimensionless (its argument must be dimensionless, and by convention, if $ t $ is in an appropriate unit where the argument is dimensionless, the function itself is dimensionless).

Therefore, $ A $ must carry the dimension of length:

$$ [A] = L. $$

Term $ B \cos^2 t $:

Similarly, the cosine function is dimensionless, and so is its square.

Thus, $ B $ must have the dimension of length:

$$ [B] = L. $$

Term $ C t^2 $:

Here, $ t^2 $ has the dimension $ T^2 $.

To have the overall term with the dimension $ L $, we require:

$$ [C] \times [T^2] = L. $$

So, the dimension of $ C $ must be:

$$ [C] = L \, T^{-2}. $$

Term $ D $:

This is a constant term added to position, so it must also have the dimension of length:

$$ [D] = L. $$

Now, we are asked to find the dimension of:

$$ \frac{ABC}{D}. $$

The dimensions of $ A $, $ B $, and $ C $ are:

$$ [A] = L, \quad [B] = L, \quad [C] = L \, T^{-2}. $$

Multiplying them together:

$$ [ABC] = L \times L \times (L \, T^{-2}) = L^3 \, T^{-2}. $$

Since $ [D] = L $, dividing by $ D $ gives:

$$ \frac{[ABC]}{[D]} = \frac{L^3 \, T^{-2}}{L} = L^2 \, T^{-2}. $$

Thus, the dimension of $ \frac{A B C}{D} $ is:

$$ \boxed{L^2 T^{-2}}. $$

The correct answer is Option C.