= Solution
The fluctuation speed $q$ and the <integral scale of turbulence> $L_u$ give a turnover time $L_u/q$. Dividing the velocity-variance scale $q^2$ by that time gives a <turbulent kinetic energy dissipation rate> proportional to $q^3/L_u$. Likewise <mass density> variance $\overline{\rho'^2}$ removed over scalar mixing time $L_\rho/q$ gives the <density-variance dissipation rate> proportional to $\overline{\rho'^2}q/L_\rho$. Absorbing dimensionless proportionality coefficients into the specified length scales gives
$$
\boxed{\epsilon=\frac{q^3}{L_u},\qquad\chi=\frac{\overline{\rho'^2}q}{L_\rho}.}
$$
These are integral-scale closure estimates, not exact consequences of dimensional analysis alone. In particular $q^2=2k$ for the turbulent-energy convention in Question 3. The normalization of $\chi$ is the one fixed by the printed scalar balance; if the budget is written for the full variance rather than one half of it, both production and dissipation terms acquire a factor two.
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