Single-phase line capacitance, 6 mm, 1.3 m

For a single-phase line with ground effect neglected, the capacitance between the conductors is $C_{ab} = \pi\varepsilon_0/\ln(D/r)$, and the capacitance of each conductor to ground (neutral) is $C_n = 2C_{ab}$, since the neutral plane lies midway and $V_{an} = V_{ab}/2$.

"Conductors of 6 mm" is read as radius 6 mm.

   a o<-------- 1.3 m -------->o b
           (neutral plane midway)

Data: $r = 0.006$ m, $D = 1.3$ m.

$$\ln\frac{D}{r} = \ln\frac{1.3}{0.006} = \ln 216.67 = 5.3784$$

Capacitance between conductors

$$\begin{aligned} C_{ab} &= \frac{\pi\varepsilon_0}{\ln(D/r)} = \frac{\pi \times 8.854\times10^{-12}}{5.3784} \\ &= 5.172\times10^{-12}\ \text{F/m} = 5.172\ \text{nF/km} \end{aligned}$$

Capacitance of each conductor to ground (neutral)

$$C_n = 2C_{ab} = \frac{2\pi\varepsilon_0}{\ln(D/r)} = 10.344\times10^{-12}\ \text{F/m} = 10.344\ \text{nF/km}$$

  a o--||--o--||--o b      C_ab = C_n/2
       C_n  |  C_n         (two C_n in series)
           ground

Answer: $C_{ab} = 5.17$ nF/km; $C_n = 10.34$ nF/km. (If 6 mm were the diameter, $r = 3$ mm, giving $C_{ab} = 4.58$ nF/km and $C_n = 9.16$ nF/km.)