C

Step 1: Define the physical quantities and find their respective dimensions.

(i) Coefficient of viscosity: The coefficient of viscosity ($\eta$) is defined as the ratio of shear stress to the shear rate. The units are given by:
$$\left[ \eta \right] = \frac{F/A}{du/dy} = \frac{MLT^{-2}}{L/T} = M L^{-1} T^{-1}$$
Now we know that the coefficient of viscosity should have dimensions $[M][L]^{-1}[T]^{-1}$, which does not match any option but helps eliminate some of the others.

(ii) Surface tension: Surface tension ($\gamma$) is defined as force per unit length, thus its dimension is:
$$\left[ \gamma \right] = \frac{F}{L} = \frac{MLT^{-2}}{L} = M T^{-2}$$
This corresponds to option [B] $ML^0T^{-2}$.

(iii) Modulus of rigidity: Modulus of rigidity (shear modulus) is given by:
$$\text{Modulus of rigidity} = \frac{F/A}{\Delta x/L} = \frac{MLT^{-2}}{L^{2}/L} = ML^{-1}T^{-2}$$
This corresponds to option [C] $ML^{-1}T^{-2}$.

(iv) Energy per unit volume of fluid: Energy per volume is defined as energy (Joules) per unit volume (m$^3$), and its dimensions are:
$$\left[ Energy \right] = [ML^{2}T^{-2}], \quad [Volume] = [L^{3}], \quad \Rightarrow \left[\frac{Energy}{Volume}\right] = \frac{ML^{2}T^{-2}}{L^{3}} = ML^{-1}T^{-2}$$
Since energy per unit volume matches with modulus of rigidity and coefficient of viscosity counts for options seen, we conclude that the matching is likely: (i) [C], (ii) [B], (iii) [A], (iv) [D].

Therefore, the correct match is (iii) Modulus of rigidity with [C].