= Solution
The <polymer contour length> and magnitude of total <electric charge> are
$$
L=20000(0.34\,\mathrm{nm})=6.80\,\mathrm{\mu m},\qquad
Q=\lambda L=40000e=6.409\times10^{-15}\,\mathrm C.
$$
Use a uniform charged-sphere model for the packed <DNA>, a homogeneous dielectric with $\epsilon=\epsilon_r\epsilon_0$, and initially neglect <electrostatic screening>. <Gauss's law> gives $E(r)=Qr/(4\pi\epsilon R^3)$ inside the sphere and $E(r)=Q/(4\pi\epsilon r^2)$ outside. Integrating the <electrostatic energy> both inside and outside,
$$
U_c=\frac\epsilon2\int E^2\,dV
=\frac{Q^2}{40\pi\epsilon R}+\frac{Q^2}{8\pi\epsilon R}
=\boxed{\frac{3Q^2}{20\pi\epsilon R}}.
$$
For an illustrative aqueous relative <permittivity> $\epsilon_r=80$ and $R=20\,\mathrm{nm}$,
$$
\boxed{U_c=1.384\times10^{-13}\,\mathrm J=1.384\times10^8\,\mathrm{pN\,nm}.}
$$
The question does not specify <permittivity>; the result scales as $80/\epsilon_r$. Using vacuum instead would increase it eightyfold and would not describe an aqueous interior.
Strictly, the <work> of packaging is a difference of <electrostatic energies>, rather than the packed self-energy alone. An extended charged rod also has self-energy. With a microscopic cutoff $a$ of order the <DNA> radius, the thin-rod approximation is
$$
U_{\rm rod}\simeq\frac{\lambda^2L}{4\pi\epsilon}\left[\ln\left(\frac{2L}{a}\right)-1\right].
$$
One derivation integrates the pair <Coulomb potential> along the rod, replacing separation by $\sqrt{z^2+a^2}$: the exact regularized expression is $\lambda^2[L\operatorname{arsinh}(L/a)-\sqrt{L^2+a^2}+a]/(4\pi\epsilon)$. The diagram's $1\,\mathrm{nm}$ transverse label suggests $a\simeq0.5\,\mathrm{nm}$, giving $U_{\rm rod}\simeq6.25\times10^6\,\mathrm{pN\,nm}$. Hence this explicit reference-state estimate is $\boxed{\Delta U\simeq1.32\times10^8\,\mathrm{pN\,nm}}$. The cutoff and precise initial conformation are additional model assumptions, but their correction is small compared with the unscreened sphere value.
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