The electric field in a region of space is given by, $$\vec{E} = E_0\hat{i} + 2E_0\hat{j}$$ where $$E_0 = 100$$ N C$$^{-1}$$. The flux of this field through a circular surface of radius 0.02 m parallel to the Y-Z plane is nearly:

The electric field is given by $$\vec{E} = E_0 \hat{i} + 2E_0 \hat{j}$$, where $$E_0 = 100$$ N C$$^{-1}$$. So, substituting the value, $$\vec{E} = 100 \hat{i} + 2 \times 100 \hat{j} = 100 \hat{i} + 200 \hat{j}$$ N C$$^{-1}$$.

The circular surface has a radius of 0.02 m and is parallel to the Y-Z plane. Since it is parallel to the Y-Z plane, the normal to the surface is along the X-axis. Therefore, the area vector $$\vec{A}$$ is directed along the $$\hat{i}$$ direction. The magnitude of the area vector is the area of the circle, which is $$A = \pi r^2$$.

Substituting the radius $$r = 0.02$$ m, we get $$A = \pi \times (0.02)^2 = \pi \times 0.0004 = 0.0004\pi$$ m$$^2$$. Thus, $$\vec{A} = 0.0004\pi \hat{i}$$ m$$^2$$.

The electric flux $$\phi$$ through the surface is given by the dot product $$\phi = \vec{E} \cdot \vec{A}$$. So, $$\phi = (100 \hat{i} + 200 \hat{j}) \cdot (0.0004\pi \hat{i})$$.

Computing the dot product, the $$\hat{i}$$ components multiply and the $$\hat{j}$$ component of $$\vec{E}$$ multiplies with the $$\hat{j}$$ component of $$\vec{A}$$, but since $$\vec{A}$$ has no $$\hat{j}$$ component, that term is zero. Therefore:

$$\phi = (100 \times 0.0004\pi) + (200 \times 0) = 100 \times 0.0004\pi = 0.04\pi$$ N m$$^2$$ C$$^{-1}$$.

Now, substituting $$\pi \approx 3.14$$, we get $$\phi = 0.04 \times 3.14 = 0.1256$$ N m$$^2$$ C$$^{-1}$$. This value is approximately 0.125 N m$$^2$$ C$$^{-1}$$.

Comparing with the options, 0.125 N m$$^2$$ C$$^{-1}$$ matches option C.

Hence, the correct answer is Option C.