Read the following information carefully and answer the questions given below. The information about three sellers who sold two types of rice, i.e., brown and white.

The quantity of white rice sold by all three seller is 230 kg. The quantity of brown rice sold by B is 20% more than the quantity of white rice sold by A. The quantity of white rice sold by B is 20 kg more than that of brown rice. The ratio of brown rice sold by A to B is 2:3. The average quantity of brown rice sold by all three sellers is 60 kg, and the total rice (white and brown) sold by C is 75 kg.

The total quantity of white rice sold by C is how much more or less than the total quantity of brown rice sold by A (in Kg).

To find the required difference, let us first define the variables for the quantity of rice sold by each seller (A, B, and C):
Let $W_A$, $W_B$, and $W_C$ be the quantity of white rice sold by A, B, and C respectively.
Let $B_A$, $B_B$, and $B_C$ be the quantity of brown rice sold by A, B, and C respectively.

From the given information, we can establish the following equations:
1. The total quantity of white rice sold by all three sellers is 230 kg:
$W_A+W_B+W_C=230$
2. The average quantity of brown rice sold by all three sellers is 60 kg, which gives the total brown rice as:
$B_A+B_B+B_C=3\times 60=180$
3. The total rice sold by C is 75 kg:
$W_C+B_C=75$
4. The quantity of brown rice sold by B is 20% more than the white rice sold by A:
$B_B=1.2\times W_A$ ⇒ $W_A=\frac{5}{6}{B}_B$
5. The ratio of brown rice sold by A to B is 2:3:
$B_A=\frac{2}{3}{B}_B$
6. The quantity of white rice sold by B is 20 kg more than that of brown rice sold by B:
$W_B=B_B+20$

Now, let us substitute the expressions of $W_A$ and $W_B$ in the first equation:
$\left(\frac{5}{6}{B}_B\right)+\left(B_B+20\right)+W_C=230$
$\frac{11}{6}{B}_B+W_C=210$ (Equation I)

Next, let us substitute the expression of $B_A$ in the second equation:
$\left(\frac{2}{3}{B}_B\right)+B_B+B_C=180$
$\frac{5}{3}{B}_B+B_C=180$ (Equation II)

Adding Equation I and Equation II together, we get:
$\left(\frac{11}{6}{B}_B+\frac{5}{3}\right)+\left(W_C+B_C\right)=210+180$
Since $W_C+B_C=75$:
$\left(\frac{11}{6}+\frac{10}{6}\right)B_B+75=390$
$\frac{21}{6}{B}_B=315$
$\frac{7}{2}{B}_B=315$
$B_B=\frac{315\times 2}{7}=90\text{kg}$

Now we can compute the required values:
- Brown rice sold by A: $B_A=\frac{2}{3}\times 90=60\text{kg}$
- Brown rice sold by C: Using Equation II, $B_C=180-\left(\frac{5}{3}\times 90\right)=180-150=30\text{kg}$
- White rice sold by C: $W_C=75-B_C=75-30=45\text{kg}$

Finally, we calculate the difference between the total quantity of white rice sold by C and brown rice sold by A:
$B_A-W_C=60-45=15\text{kg}$

Thus, the total quantity of white rice sold by C is 15 kg less than the total quantity of brown rice sold by A.