1. Understand the geometry

A uniformly charged metallic ring has two perpendicular diameters $AC$ and $BD$. So the points $A,B,C,D$ divide the circle into four equal arcs. Hence:

• arc $AB$ is a quarter circle,
• arc $BC$ is also a quarter circle,
• arc $ABC$ is a semicircle.

We are given that the electric field at the center $O$ due to arc $AB$ has magnitude $E$. We need the electric field at $O$ due to arc $ABC$.

2. Relation between fields of arcs $AB$, $BC$, and $ABC$

Because the charge distribution is uniform and arcs $AB$ and $BC$ are equal in length, the magnitudes of electric fields at $O$ due to these two arcs are equal.

So, |ec E_{AB}| = |ec E_{BC}| = E.

Now we determine their directions.

3. Direction of field due to arc $AB$

For a positively charged arc, the electric field at the center is along the bisector of the arc, directed away from the arc.

• Arc $AB$ lies in the first quadrant.
• Its bisector makes $45^\circ$ with the axes.
• Therefore, $\vec E_{AB}$ points toward the third quadrant along the line making $225^\circ$ from the positive $x$-axis.

Similarly:

• Arc $BC$ lies in the second quadrant.
• Its field $\vec E_{BC}$ points toward the fourth quadrant, along the bisector of that arc.

Thus the angle between $\vec E_{AB}$ and $\vec E_{BC}$ is $90^\circ$.

4. Add the two field vectors

The field due to arc $ABC$ is the vector sum: $\vec E_{ABC} = \vec E_{AB} + \vec E_{BC}.$

Since both have magnitude $E$ and are perpendicular, $|\vec E_{ABC}| = \sqrt{E^2 + E^2} = \sqrt{2}\,E.$

5. Final answer

Therefore, the magnitude of electric field at $O$ due to arc $ABC$ is $\boxed{\sqrt{2}E}.$

So the correct option is D.

6. Comparison with stored answer

Stored correct answer: D

Our derived answer: D

They agree.