Step 1: Recall the relationship between electric field and potential.
The potential difference between two points A and B is:
\[
V(B) - V(A) = -\int_A^B \vec{E} \cdot d\vec{l}
\]
Step 2: Choose a convenient path from origin to point (10, 20).
We integrate in two steps: first from $(0,0)$ to $(10,0)$ along the $x$-axis, then from $(10,0)$ to $(10,20)$ along the $y$-direction. This avoids cross terms.
Step 3: Integrate along the x-axis from (0,0) to (10,0).
On this path $y = 0$ and $d\vec{l} = dx\,\hat{i}$, so $\vec{E} \cdot d\vec{l} = Ax\,dx$:
\[
V(10,0) - V(0,0) = -\int_0^{10} Ax\,dx = -A\left[\frac{x^2}{2}\right]_0^{10} = -\frac{A \times 100}{2} = -50A
\]
Step 4: Integrate along y from (10,0) to (10,20).
On this path $x = 10$ and $d\vec{l} = dy\,\hat{j}$, so $\vec{E} \cdot d\vec{l} = Ay\,dy$:
\[
V(10,20) - V(10,0) = -\int_0^{20} Ay\,dy = -A\left[\frac{y^2}{2}\right]_0^{20} = -\frac{A \times 400}{2} = -200A
\]
Step 5: Combine the two steps.
\[
V(10,20) - V(0,0) = -50A - 200A = -250A
\]
With $A = 10\,\text{V\,m}^{-2}$:
\[
V(10,20) - V(0,0) = -250 \times 10 = -2500\,\text{V}
\]
Step 6: Use the condition that V(10,20) = 0 to find V(0,0).
\[
0 - V(0,0) = -2500 \implies V(0,0) = 2500\,\text{V}
\]
\[
\boxed{2500\,\text{V}}
\]