To solve this problem, we need to use the formulas and concepts related to parallel resonant circuits. Let's break down the problem into smaller steps to find the value of capacitance (C).

1. Understanding the given information:
- Midband admittance (Ym) = 25 x 10^(-3) S (Siemens)
- Quality factor (Q) = 80
- Resonant frequency (fr) = 200 krad/s

2. Relationships in a parallel resonant circuit:
In a parallel resonant circuit, the admittance (Y) is given by the equation:
Y = Ym + j(Bw - Gw)

Where:
- Y is the total admittance
- Ym is the midband admittance
- Bw is the susceptance due to reactive components
- Gw is the conductance due to resistive components
- j is the imaginary unit (j^2 = -1)
- w is the angular frequency (w = 2πf, where f is the frequency)

3. Calculating the susceptance (Bw):
Since the circuit is parallel resonant, at resonance, the susceptance is zero (Bw = 0). This means that the circuit behaves as a pure resistance at resonance.

4. Finding the conductance (Gw):
The conductance (Gw) can be calculated using the formula:
Gw = wC

5. Calculating the resonant frequency (fr) in radians/second:
Given the resonant frequency (fr) = 200 krad/s, we can directly use this value.

6. Solving for the conductance (Gw):
Using the formula Gw = wC, we can rearrange it to solve for C:
C = Gw / w

7. Calculating the value of C:
Substituting the values of Gw = wC and Bw = 0 into the admittance equation, we get:
Y = Ym + j(0 - Gw)
Y = Ym - jGw

At resonance, the circuit behaves as a pure resistance, so the imaginary component of the admittance (jGw) is zero. Therefore, the admittance (Y) becomes equal to the midband admittance (Ym).

Y = Ym
Ym = 25 x 10^(-3) S

Now, substituting the values of Ym and Gw into the equation C = Gw / w, we get:
C = Gw / w
C = (200 x 10^3) / (2π x 200 x 10^3)
C = 1 / (2π)
C ≈ 0.159 F

The closest option to the calculated value of C is 10 F (option C).

Therefore, the correct answer is option C) 10 F.