A quantity X is given by ε₀L( ΔV/Δt) where ε₀ is the the permittivity of the free space, L is a length, ΔV is a potential difference and Δt is a time interval. The dimensional formula for X is the same as that of

The correct option is current.

To find the physical quantity that has the same dimensional formula as $X$, let us analyze the given formula:
$X={\epsilon }_{0}L\left(\frac{\Delta V}{\Delta t}\right)$
where ${\epsilon }_0$ is the permittivity of free space, $L$ is a length, $\Delta V$ is a potential difference, and $\Delta t$ is a time interval.

First, recall the formula for the capacitance $C$ of a parallel-plate capacitor:
$C=\frac{{\epsilon }_{0}A}{d}$
Here, $A$ represents area (dimension of $L^2$) and $d$ represents distance (dimension of $L$).

In terms of dimensions, we can write:
$\left[C\right]=\frac{\left[{\epsilon }_0\right]·\left[L^2\right]}{\left[L\right]}=\left[{\epsilon }_0\right]\left[L\right]$
This shows that the product ${\epsilon }_{0}L$ has the dimensions of capacitance $C$.

Now, substituting capacitance $C$ in place of ${\epsilon }_{0}L$ in our equation for $X$, we get:
$X=C\left(\frac{\Delta V}{\Delta t}\right)$

Since the charge $Q$ on a capacitor is given by the relation $Q=CV$, the term $C·\Delta V$ represents the change in charge $\Delta Q$.

Substituting this relation into the equation, we obtain:
$X=\frac{\Delta Q}{\Delta t}$

The quantity $\frac{\Delta Q}{\Delta t}$ represents the rate of flow of charge with time, which is the definition of electric current. Therefore, the dimensional formula of $X$ is the same as that of current.