= Solution
A freely swimming body has no net external force or torque in <Stokes flow>. Its <force-free> condition removes the $r^{-1}$ <Stokeslet> monopole, and its <torque-free> condition removes the antisymmetric force-dipole contribution. The leading generic far field is therefore a <symmetric traceless rank-two tensor> <stresslet>, decaying as $r^{-2}$. Its axial strength may vanish for special swimmers, in which case higher-order singularities dominate.
The <axisymmetric stresslet from a Stokeslet pair> provides an explicit construction and fixes the sign convention. Place forces $+F\mathbf e$ and $-F\mathbf e$ on the fluid at $+(a/2)\mathbf e$ and $-(a/2)\mathbf e$. The <Stokeslet> tensor is
$$
J_{ij}(\mathbf r)=\frac1{8\pi\mu}
\left(\frac{\delta_{ij}}r+\frac{r_i r_j}{r^3}\right).
$$
Expanding their sum for $r\gg a$ gives
$$
\mathbf u
=FJ(\mathbf r-\tfrac a2\mathbf e)\mathbf e
-FJ(\mathbf r+\tfrac a2\mathbf e)\mathbf e
=-Fa\,(\mathbf e\cdot\nabla)[J(\mathbf r)\mathbf e]+\cdots.
$$
Since
$$
(\mathbf e\cdot\nabla)[J(\mathbf r)\mathbf e]
=\frac1{8\pi\mu}
\left[\frac{\mathbf r}{r^3}
-\frac{3(\mathbf e\cdot\mathbf r)^2\mathbf r}{r^5}\right],
$$
the dipole strength is $S=Fa$ and the resulting <stresslet> is the given one. More general surface-traction distributions produce the same leading form through their symmetric traceless first force moment.
Writing $\cos\alpha=\mathbf e\cdot\widehat{\mathbf r}$, its radial velocity is
$$
u_r=\frac{S}{8\pi\mu r^2}(3\cos^2\alpha-1).
$$
Consequently
$$
\boxed{S>0:\ \text{pusher, axial outflow and equatorial inflow};
\qquad
S<0:\ \text{puller, axial inflow and equatorial outflow}.}
$$
This classifies <pusher microswimmers> and <puller microswimmers> in the force-on-fluid convention used by the given formula.
For the circular swimmer let $\varphi=\omega t$, $\mathbf p=(\cos\varphi,\sin\varphi,0)$, $\mathbf x'=R\mathbf p$, and choose the tangent director $\mathbf e=(-\sin\varphi,\cos\varphi,0)$. Its sign is immaterial to the <stresslet>. Write $\mathbf r=\mathbf x$ for the observation vector from the circle centre, and assume $r\gg R$. The leading <far-field orbit average of a tangent stresslet> is obtained by replacing $\mathbf x-\mathbf x'$ by $\mathbf r$ while retaining the time-dependent director.
The <orientation averaging of an axisymmetric stresslet> over a full revolution gives
$$
\langle e_i e_j\rangle
=\frac12(\delta_{ij}-\widehat z_i\widehat z_j),\qquad
\langle(\mathbf e\cdot\mathbf r)^2\rangle
=\frac12(r^2-z^2).
$$
Substitution gives
$$
\begin{aligned}
\langle\mathbf u(\mathbf r)\rangle
&=\frac{S}{8\pi\mu}
\left[-\frac{\mathbf r}{r^3}
+\frac32\frac{(r^2-z^2)\mathbf r}{r^5}\right]
+\text{higher multipoles}\\
&=-\frac{S/2}{8\pi\mu}
\left[-\frac{\mathbf r}{r^3}
+\frac{3z^2\mathbf r}{r^5}\right]
+\text{higher multipoles}.
\end{aligned}
$$
Thus
$$
\boxed{\widetilde S=-\frac S2,\qquad
\widetilde{\mathbf e}=\widehat{\mathbf z}}
$$
with $\widetilde{\mathbf e}$ equivalently $-\widehat{\mathbf z}$. The effective axis is normal to the orbit plane. A positive original strength becomes negative, and a negative strength becomes positive: \b[a pusher averages to a puller, and a puller averages to a pusher], at leading far-field order. The result is independent of $R$ and $\omega$ at this order.
The accuracy of the centre replacement can be made explicit. Half a revolution changes $\mathbf p$ and $\mathbf e$ to their negatives; the <stresslet> is even in $\mathbf e$. Pairing those times replaces the displaced field by
$$
\frac12\left[G(\mathbf r-R\mathbf p,\mathbf e)
+G(\mathbf r+R\mathbf p,\mathbf e)\right]
=G(\mathbf r,\mathbf e)+O\!\left(\frac{|S|R^2}{\mu r^4}\right).
$$
Hence
$$
\boxed{\langle\mathbf u\rangle
=G(\mathbf r;\widetilde S,\widehat{\mathbf z})
+O\!\left(\frac{|S|R^2}{\mu r^4}\right).}
$$
The finite-radius average is not exactly a point <stresslet> at every location. For example, on the positive $z$ axis the exact average of the supplied model is
$$
\langle\mathbf u(0,0,z)\rangle
=-\frac{S}{8\pi\mu}\frac{z}{(z^2+R^2)^{3/2}}\,\widehat{\mathbf z},
$$
which has the derived point-stresslet limit for $z\gg R$. This <even displacement correction in an orbit-averaged stresslet> is consistent with the far-field restriction.
\Image[/past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2015/iii/paper-80-stresslet-average.png]
{title=Instantaneous pusher along an in-plane axis and its leading circular-orbit average, a puller normal to the orbit plane}
{height=440}
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