According to Eq. 4.5, the force on a single dipole is
\[\mathbf{F}=(\mathbf{p}\cdot \nabla )\mathbf{E} \tag{Eq. 4.5, page 170}\]So the net force on a dielectric object is
\[\mathbf{F}=\int (\mathbf{P}\cdot \nabla )\mathbf{E}_{\text{ext} }d\tau \tag{Eq. 4.69} \]Here $\mathbf{E}_{\text{ext} } $) is the field of everything except the dielectric.
You might assume that it wouldn't matter if you used the total field, after all, the dielectric can't exert a force on itself.
However, because the field of the dielectric is discontinuous at the location of any bound surface charge, the derivative induces a spurious delta function, and it is safer to stick to $\mathbf{E}_{\text{ext} } $.
Use Eq. 4.69 to determine the force on a tiny sphere, of radius $R $, composed of linear dielectric material of susceptibility $\chi _e $, which is situated a distance $s $ from a fine wire carrying a uniform line charge $\lambda $.
In this scenario, the external field is that of the uniform line charge. The electric field of such line charge points in its cylindrical radial direction:
\[\mathbf{E}_{\text{ext} }=\frac{\lambda }{2\pi \epsilon _0 }\frac{\hat{\mathbf{s}}}{s} \]In this solution I will assume that this field is roughly constant around the tiny sphere, so for practical purposes we can consider the scenario presented in Example 4.7 (page 193). In that example, after using boundary conditions to solve Laplace's equation, we find that the field inside the sphere is
\[\mathbf{E}_{\text{in} }=\frac{3}{\epsilon _r+2}\mathbf{E}_0 \]Where $\mathbf{E}_0 $ denotes the constant field outside. Now that we know the $\mathbf{E} $ field, we can compute $\mathbf{P} $:
\begin{align*} \mathbf{P} &= \epsilon _0 \chi _e\mathbf{E}_0\tag{Eq. 4.30, page 186} \\ &= \epsilon _0 \chi _e \frac{3}{\epsilon _r+2}\mathbf{E}_0\\ &= \epsilon _0 \chi _e \frac{3}{1+\chi _e+2}\mathbf{E}_0\\ &= \epsilon _0 \chi _e \frac{3}{3+\chi _e}\mathbf{E}_0\\ &= \epsilon _0 \chi _e \frac{1}{1+\frac{\chi _e}{3}}\mathbf{E}_0 \end{align*}In the above, we will replace $\mathbf{E}_0 $ by $\mathbf{E}_{\text{ext} } $. Both represent the same thing, but it's important to keep notation consistent!
\[\mathbf{P}= \epsilon _0 \chi _e \frac{1}{1+\frac{\chi _e}{3}}\mathbf{E}_{\text{ext} }\]We are now ready to use Equation 4.69:
\begin{align*} \mathbf{F} &= \int (\mathbf{P}\cdot \nabla )\mathbf{E}_{\text{ext} }d\tau \\ &= \int \left( \epsilon _0 \chi _e \frac{1}{1+\frac{\chi _e}{3}}\mathbf{E}_{\text{ext} }\cdot \nabla \right) \mathbf{E}_{\text{ext} }d\tau \\ &= \left( \epsilon _0 \chi _e \frac{1}{1+\frac{\chi _e}{3}} \right) \frac{\lambda }{2\pi \epsilon _0 s}\frac{\partial }{\partial s}\frac{\lambda }{2\pi \epsilon _0 }\frac{\hat{\mathbf{s}}}{s}\int d\tau\\ &= -\frac{\chi _e}{1+\frac{\chi _e}{3}}\frac{\lambda ^2 }{4\pi ^2 \epsilon _0 }\frac{1}{s^2 }\hat{\mathbf{s}}\int d\tau \\ &= -\frac{\chi _e}{1+\frac{\chi _e}{3}}\frac{\lambda ^2 }{4\pi ^2 \epsilon _0 }\frac{1}{s^2 } \left( \frac{4}{3}\pi R^3 \right) \hat{\mathbf{s}} \end{align*}Where I took $s $'s out of the integral since we assume that the sphere is tiny, so $s $ won't vary much as we integrate over the sphere. Further, the integral $\int d\tau $ was performed over the tiny sphere of radius $R $.