ML³ T⁻¹Q² is dimension of

The correct option is Resistivity .

Let's derive the dimensional formula step-by-step to identify the quantity that matches ${\mathrm{ML}}^3{T}^{-1}{Q}^{-2}$ (or ${\mathrm{ML}}^3{T}^{-1}{Q}^2$ depending on charge notation, let's analyze carefully with respect to charge $Q$).
Recall the relation between resistance $R$ and resistivity $\rho$:
$R=\rho \frac{l}{A}$
where $l$ is length and $A$ is area. Thus, resistivity $\rho$ is given by:
$\rho =R\frac{A}{l}$

First, let's find the dimensions of resistance $R$ using Ohm's Law ($V=IR$):
$R=\frac{V}{I}$
Since potential difference $V$ is work done ($W$) per unit charge ($Q$):
$V=\frac{W}{Q}$
The dimensional formula of work $W$ is $\left[{\mathrm{ML}}^2{T}^{-2}\right]$. Therefore, the dimension of potential $V$ in terms of charge $Q$ is:
$\left[V\right]=\frac{\left[{\mathrm{ML}}^2{T}^{-2}\right]}{\left[Q\right]}=\left[{\mathrm{ML}}^2{T}^{-2}{Q}^{-1}\right]$

Electric current $I$ is rate of flow of charge:
$I=\frac{Q}{T}$
So, the dimension of current $I$ is:
$\left[I\right]=\left[{\mathrm{QT}}^{-1}\right]$

Now, substituting the dimensions of $V$ and $I$ to find the dimensions of resistance $R$:
$\left[R\right]=\frac{\left[{\mathrm{ML}}^2{T}^{-2}{Q}^{-1}\right]}{\left[{\mathrm{QT}}^{-1}\right]}=\left[{\mathrm{ML}}^2{T}^{-1}{Q}^{-2}\right]$

Now we calculate the dimension of resistivity $\rho$:
$\left[\rho \right]=\left[R\right]\times \frac{\left[A\right]}{\left[l\right]}$
Since area $\left[A\right]=\left[L^2\right]$ and length $\left[l\right]=\left[L\right]$:
$\left[\rho \right]=\left[{\mathrm{ML}}^2{T}^{-1}{Q}^{-2}\right]\times \frac{\left[L^2\right]}{\left[L\right]}$
$\left[\rho \right]=\left[{\mathrm{ML}}^3{T}^{-1}{Q}^{-2}\right]$

This matches the given dimensional formula (taking the standard notation where charge $Q$ is represented). Hence, the quantity is Resistivity.