= Solution
For a static finite-energy <field>, the <Hamiltonian density> is $\tfrac12(\phi')^2+V(\phi)$. Multiply the static <equation of motion> by $\phi'$ and integrate once:
$$
\frac{d}{dx}\left[\frac12(\phi')^2-V(\phi)\right]=0.
$$
At either vacuum endpoint both terms vanish, so $\tfrac12(\phi')^2=V$. The total <energy density> is consequently $(\phi')^2=4k^2\operatorname{sech}^2[k(x-x_0)]/b^2$. Its integral gives
$$
\boxed{E_0=\frac{4k^2}{b^2}\frac2k=\frac{8\sqrt{ab}}{b^2}.}
$$
For the usual positive $a,b$ convention this is
$$
\boxed{E_0=c\sqrt a,\qquad c=\frac8{b^{3/2}}.}
$$
The coefficient depends on the dimensionless parameter $b$; it is not a universal numerical constant independent of the potential. In two <spacetime> dimensions, $[\phi]=[b]=0$ and $[a]=2$, so the dimensions are consistent. If both $a$ and $b$ are negative while $ab>0$, the real answer is instead $8\sqrt{|a|}/|b|^{3/2}$; the literal factor $\sqrt a$ would not be real.
An equivalent positive-parameter calculation substitutes $z=b\phi$ in $E_0=\int_0^{2\pi/b}\sqrt{2V}\,d\phi$ and uses the provided integral of $\sqrt{1-\cos z}$. The following original plot shows that the localized <energy density> sits where the <field> passes between the vacua.
\Image[/past-exam-of-the-theoretical-physics-major-option-course-of-the-university-of-oxford/2016-sine-gordon-kink-energy.png]
{title=The dimensionless sine-Gordon kink profile and its localized static energy density, with total dimensionless energy eight}
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