**Solution:**

Given data:
- Voltage amplitude (V) = 120 V
- Voltage angle (θ) = 0°
- Frequency (f) = 400 rad/sec
- Current angle (φ) = 63.5°
- Inductance (L) = 25 mH
- Capacitance (C) = 50 μF

To find the value of resistance (R), we can use the relationship between resistance, inductance, and capacitance in an RLC circuit. The impedance (Z) of the circuit is given by:

Z = √(R^2 + (ωL - 1/ωC)^2)

Where:
- ω = 2πf (angular frequency)

To find the value of R, we need to solve the equation for impedance. Since we are given the voltage and current angles, we can use the relationship between impedance and the phase angle difference (Δθ) to solve for R. The phase angle difference is given by:

Δθ = θ - φ

Where:
- θ = Voltage angle
- φ = Current angle

Let's calculate the phase angle difference:

Δθ = 0° - 63.5°
Δθ = -63.5°

Now, we can solve for R:

Z = V / I
Z = V / (V / Z)
Z = 1

1 = √(R^2 + (ωL - 1/ωC)^2)
1 = √(R^2 + ((2πf)L - 1/(2πf)C)^2)
1 = √(R^2 + (2πfL - 1/(2πfC))^2)
1 = √(R^2 + (2π(400)(0.025) - 1/(2π(400)(50*10^-6)))^2)

Now, we can solve for R by substituting the values and solving the equation:

1 = √(R^2 + (100π - 1/(800π))^2)
1 = √(R^2 + (314.16 - 1/(502.65))^2)
1 = √(R^2 + (314.16 - 0.00199)^2)
1 = √(R^2 + (314.16 - 0.00199)^2)
1 = √(R^2 + 314.15801^2)
1 = √(R^2 + 98648.7471)
1 = R^2 + 98648.7471
R^2 = 1 - 98648.7471
R^2 = -98647.7471
R = √-98647.7471
R ≈ 0 (Since the square root of a negative number is imaginary)

Therefore, the value of resistance (R) in the series RLC circuit is approximately 0.

Now, let's calculate the drops across the inductor (VL) and capacitor (VC) using the formulas:

VL = IXL
VC = IXC

Where:
- VL = Voltage drop across the inductor
- VC = Voltage drop across the capacitor
- I = Current amplitude
- XL = Inductive reactance
- XC =