= Solution
Let a segment of fixed length $b$ make angle $\theta$ with an applied <force> $\mathbf f=f\hat{\mathbf z}$. Its <energy> contribution is $-fb\cos\theta$. Independent freely jointed orientations give the single-segment <partition function> and chain <partition function>
$$
Z_1=\int e^{u\cos\theta}d\Omega=4\pi\frac{\sinh u}{u},\qquad
u=\frac{fb}{k_BT},\qquad Z_N=Z_1^N.
$$
The mean end-to-end projection is the <force> <derivative> of $k_BT\ln Z_N$:
$$
\langle R_z\rangle=Nb\left(\coth u-\frac1u\right).
$$
For $|u|\ll1$, the <Langevin function> is $u/3+O(u^3)$. <Isotropy> therefore gives the vector response
$$
\boxed{\langle\mathbf R\rangle=\frac{Nb^2}{3k_BT}\mathbf f+O(f^3),\qquad
\mathbf f=\frac{3k_BT}{Nb^2}\langle\mathbf R\rangle\quad\text{to linear order}.}
$$
The <covariance> at zero <force> provides an independent route: $\langle R_iR_j\rangle_0=Nb^2\delta_{ij}/3$, and the <Boltzmann weight>'s first-order term yields $\langle R_i\rangle=f_j\langle R_iR_j\rangle_0/(k_BT)$. This is a relation for the <mean vector> or its projection. The zero-<force> <mean vector> vanishes even though the chain has finite <root mean square> size $b\sqrt N$.
The controlled weak-<force> condition is $fb\ll k_BT$, not merely that a short part of a measured curve looks straight. With the supplied $b$, $k_BT/b\simeq10.3\,\mathrm{pN}$ at $25^\circ\mathrm C$. A raw <force> near $30\,\mathrm{pN}$, as in the figure's labelled low-extension portion, would give $u\simeq2.9$ and is not asymptotically weak for those literal segment parameters. A <force> offset or different effective model might change that interpretation, but no such correction should be invented from the plot.
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