= Solution
An eddy at height $z$ cannot extend arbitrarily far through the ground, so its largest local vertical scale is proportional to $z$. Choose $\ell\sim z$ and absorb its constant of proportionality into the coefficient. With the prescribed constant turbulent <velocity> scale, write
$$
\nu_T=Cz,\qquad C=\alpha\hat u>0.
$$
The <mixing-length closure> is $-\overline{u'w'}=\nu_TU_z$. Thus the full kinematic stress is
$$
T\equiv\frac{\tau_d}{\rho}=(\nu+\nu_T)U_z.
$$
For $z\gg\delta$, neglect the molecular term. With no internal momentum source, $T$ is constant, so
$$
U_z=\frac{T}{Cz}.
$$
Integrating between two heights gives
$$
\boxed{U(z)=U(z_r)+\frac{T}{\alpha\hat u}\log\frac{z}{z_r}.}
$$
Equivalently define an extrapolated <roughness length> $z_0$ and write $U=(T/C)\log(z/z_0)$. If $C=\kappa u_*$, the familiar <law of the wall> is $U/u_*=(1/\kappa)\log(z/z_0)$. The logarithmic profile is valid outside the inner viscous region and below scales where the constant-stress or wall-distance assumptions fail. It cannot be continued to $z=0$ to impose the smooth-wall no-slip condition.
The part b notation $\tau=\nu_TU_z$ must therefore mean $\tau=T$, a stress divided by <density>. Using the physical stress from part a on that right-hand side would omit a factor of $\rho$.
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