Assume that an electric field $$\vec{E} = 30x^2\hat{i}$$ exists in space. Then the potential difference $$V_A - V_O$$, where $$V_O$$ is the potential at the origin and $$V_A$$ the potential at x = 2 m is:

We start with the fundamental electrostatic relation that connects the electric field to the electric potential:

$$\vec{E} = -\,\nabla V$$

Along the x-axis only, this gradient reduces to a simple derivative, so the x-component of the field satisfies

$$E_x = -\,\frac{dV}{dx}.$$

The problem gives the field as $$\vec{E} = 30x^2\,\hat i,$$ which means

$$E_x = 30x^2.$$

Equating the two expressions for $$E_x$$ we have

$$30x^2 = -\,\frac{dV}{dx}.$$

Now we isolate $$dV$$:

$$\frac{dV}{dx} = -30x^2 \quad\Longrightarrow\quad dV = -30x^2\,dx.$$

To obtain the potential difference between the point $$A$$ at $$x = 2\ \text{m}$$ and the origin $$O$$ at $$x = 0\ \text{m},$$ we integrate $$dV$$ from 0 to 2:

$$V_A - V_O \;=\; \int_{x=0}^{x=2} dV \;=\; \int_{0}^{2} \left(-30x^2\right)dx.$$

We compute the integral step by step:

First, integrate $$x^2$$:

$$\int x^2\,dx = \frac{x^3}{3}.$$

Now include the constant $$-30$$ and apply the limits:

$$V_A - V_O = -30 \left[ \frac{x^3}{3} \right]_{0}^{2}.$$

Simplify the coefficient:

$$-30 \times \frac{1}{3} = -10,$$

so

$$V_A - V_O = -10 \left[x^3\right]_{0}^{2}.$$

Evaluate at the upper and lower limits:

At $$x = 2,$$ $$x^3 = 2^3 = 8.$$

At $$x = 0,$$ $$x^3 = 0^3 = 0.$$

So

$$V_A - V_O = -10 \left(8 - 0\right) = -10 \times 8 = -80.$$

The SI unit of potential difference is volt, equivalent to $$\text{J C}^{-1}.$$ Thus

$$V_A - V_O = -80\ \text{J C}^{-1}.$$

The negative sign indicates that point $$A$$ is at a lower potential than the origin.

Hence, the correct answer is Option C.