A

To find the electric potential at a point in an electric field, we can use the following relation:

$$ V = V_0 - \int_{C} \mathbf{E} \cdot d\mathbf{l} $$

where:
- $V$ is the potential at point (3m, 3m),
- $V_0$ is the potential at the origin (0,0), which is given as 0.
- $\mathbf{E}$ is the electric field vector.
- $d\mathbf{l}$ is the differential path element along the path of integration.

Given the electric field:
$$ \mathbf{E} = (10\hat{i} + 20\hat{j}) \text{ N/C} $$
And we need to move from the origin (0, 0) to the point (3m, 3m). We can take a straight-line path for simplicity. The path can be represented parametrically as:
$$(x, y) = (t, t)$$ where $t$ goes from 0 to 3.

Now, the differential path element $d\mathbf{l}$ is:
$$ d\mathbf{l} = (dx, dy) = (dt, dt) = \hat{i} dt + \hat{j} dt = (1\hat{i} + 1\hat{j}) dt $$

Now we calculate the integral:

$$ V = 0 - \int_{0}^{3} \mathbf{E} \cdot d\mathbf{l} = -\int_{0}^{3} (10\hat{i} + 20\hat{j}) \cdot (1\hat{i} + 1\hat{j}) dt $$

Calculating the dot product:
$$ \mathbf{E} \cdot d\mathbf{l} = 10(1) + 20(1) = 30 $$

Now we have:
$$ V = -\int_{0}^{3} 30 dt = -30t \bigg|_{0}^{3} = -30(3) + 30(0) = -90 $$

Thus, the electric potential at point (3m, 3m) is V = -90 V.