Given Information:
- The annular surface has radii of 1 cm and 3 cm.
- The surface carries a nonuniform surface charge density of σ = 5 nC/m^2.
- The point P is located at (0, 0, 2 cm).

To find:
- The electric potential (V) at point P.

Solution:
1. Electric Potential due to a Charged Annular Surface:
The electric potential at a point due to a charged annular surface can be calculated using the formula:

V = ∫[k * σ * ds / r]

where:
- V is the electric potential at the point.
- k is the electrostatic constant (k = 9 × 10^9 Nm^2/C^2).
- σ is the surface charge density.
- ds is the differential area element on the annular surface.
- r is the distance between the differential area element and the point where the potential is being calculated.

2. Differential Area Element:
The differential area element (ds) on the annular surface can be calculated using the formula:

ds = 2πr * dr

where:
- ds is the differential area element.
- r is the radius of the annular surface.
- dr is the differential radius.

3. Calculating the Electric Potential at Point P:
In this case, the annular surface has radii of 1 cm and 3 cm. Let's calculate the electric potential at point P(0, 0, 2 cm) using the given values:

- The surface charge density (σ) = 5 nC/m^2.
- The distance between the annular surface and point P (r) = 2 cm = 0.02 m.

Using the formulas mentioned above, we can calculate the electric potential (V) at point P:

V = ∫[k * σ * ds / r]
= ∫[k * σ * 2πr * dr / r]
= k * σ * 2π * ∫[dr]
= k * σ * 2π * (r2 - r1)
= (9 × 10^9 Nm^2/C^2) * (5 × 10^-9 C/m^2) * 2π * (0.03 - 0.01)
= 6.78 × 10^2 V
≈ 680 mV
≈ 68 V

Therefore, the electric potential at point P is approximately 680 mV, which corresponds to option (A) in the given answer choices.