= Solution
The thin-layer condition is
$$
\boxed{C=(\nu^2/F)^{1/3}\ll1,\quad\text{equivalently }F\gg\nu^2.}
$$
Integrate the differential equation once: $(f''+ff')'=0$. Since $f(0)=f''(0)=0$, its integration constant is zero. A second integration gives $f'+f^2/2=b$. The outward-jet solution with $b>0$ and $f(0)=0$ is
$$
f=2k\tanh(k\eta),\qquad f'=2k^2\operatorname{sech}^2(k\eta),\qquad b=2k^2.
$$
Using the given fourth-power integral in the normalization gives $1=4k^4[4/(3k)]=16k^3/3$, so $k=(3/16)^{1/3}$. Thus the <similarity profile of an axisymmetric radial viscous free jet> is
$$
\boxed{u_r(r,z)=\frac{(9F^2/(32\nu))^{1/3}}r\,
\operatorname{sech}^2\left[(3F/(16\nu^2))^{1/3}\frac zr\right].}
$$
It is symmetric, outward, decays in $z$, and has exactly the conserved flux $F$. It describes the region beyond the small injection slit where the stated boundary-layer approximation applies; its formal $r=0$ singularity is outside that approximation.
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