A long coaxial cable carries a high voltage in a laboratory…

Step 1. Capacitance (a). With charge $\lambda$ per metre, $V = \dfrac{\lambda}{2\pi\varepsilon_0}\ln\dfrac{b}{a}$, so $\dfrac{C}{L} = \dfrac{2\pi\varepsilon_0}{\ln(b/a)} = \dfrac{2\pi\varepsilon_0}{\ln 4} = 40.1\ \text{pF/m}$.

Step 2. Highest voltage (b). The field at the wire is $E(a) = \dfrac{V}{a\ln(b/a)}$. Setting it to $E_b$: $V = E_b\,a\ln\dfrac{b}{a} = 3.0 \times 10^{6}\ \text{V/m} \times 0.005\ \text{m} \times 1.386 = 20.8\ \text{kV}$.

Step 3. Half the energy (c). The energy density is $\tfrac12\varepsilon_0E^2 \propto 1/r^2$, so the energy between $a$ and $r$ grows as $\ln(r/a)$. Half lies within $r$ when $\ln(r/a) = \tfrac12\ln(b/a)$: $r = \sqrt{ab} = 1.00\ \text{cm}$.

Step 4. The best wire (d). $\dfrac{d}{da}\left(a\ln\dfrac{b}{a}\right) = \ln\dfrac{b}{a} - 1 = 0$, so $a = \dfrac{b}{e} = 0.736\ \text{cm}$. A thinner wire has too strong a field at its surface; a thicker one leaves too thin a gap.

Step 5. That voltage (e). Now $\ln(b/a) = 1$, so $V = E_b\,\dfrac{b}{e} = 22.1\ \text{kV}$.

Step 6. Energy (f). $\dfrac{C}{L} = 2\pi\varepsilon_0$ now, and $\dfrac{U}{L} = \tfrac12 \cdot 2\pi\varepsilon_0V^2 = 13.5\ \text{mJ/m}$.