To determine the correct representation, we must find the option that possesses the dimensions of Time $[T]$.

1. Analyze Dimensions of Physical Quantities : Mass ($m$): $[M]$ Radius ($r$): $[L]$ Coefficient of Viscosity ($\eta$): $[ML^{-1}T^{-1}]$ Velocity ($v$): $[LT^{-1}]$

• Acceleration due to gravity ($g$): $[LT^{-2}]$

2. Analyze the Options:

• Option A ($\frac{mr^2}{6\pi\eta}$): $\frac{[M][L^2]}{[ML^{-1}T^{-1}]} = [L^3T]$. This represents volume $\times$ time, not time.
• Option B ($\sqrt{6\pi mr\eta g^2}$): $\sqrt{[M][L][ML^{-1}T^{-1}][L^2T^{-4}]} = \sqrt{[M^2 L^2 T^{-5}]}$. This does not simplify to $[T]$.
• Option C ($\frac{m}{6\pi\eta rv}$): $\frac{[M]}{[ML^{-1}T^{-1}][L][LT^{-1}]} = \frac{[M]}{[MLT^{-2}]} = [L^{-1}T^2]$. This is incorrect.

3. Theoretical Verification: The equation of motion for a particle falling under Stokes' drag is $m\frac{dv}{dt} = mg - 6\pi\eta rv$. Rearranging for the time constant (coefficient of $v$ in the differential equation), we get $\tau = \frac{m}{6\pi\eta r}$.

• Checking dimensions of $\frac{m}{6\pi\eta r}$: $\frac{[M]}{[ML^{-1}T^{-1}][L]} = \frac{[M]}{[MT^{-1}]} = [T]$.

Since the correct expression $\frac{m}{6\pi\eta r}$ is not listed among the first three options, the correct choice is 'None of the above'.