If , E and t represent the free space permittivity, electric field...

We need to find the unit of $\frac{\epsilon E}{t}$.
From Maxwell's equations, the displacement current density is given by $J_d = \epsilon \frac{\partial E}{\partial t}$.
The term $\frac{\epsilon E}{t}$ has the same dimensions as displacement current density $J_d$.
Current density $J$ is defined as current per unit area, $J = \frac{I}{A}$.
The SI unit of current $I$ is Ampere (A) and the SI unit of area $A$ is $m^2$.
Therefore, the unit of current density is $A/m^2$.
Alternatively, using dimensions:
$[\epsilon] = M^{-1}L^{-3}T^4A^2$
$[E] = MLT^{-3}A^{-1}$
$[t] = T$
Dimension of $\frac{\epsilon E}{t} = \frac{(M^{-1}L^{-3}T^4A^2)(MLT^{-3}A^{-1})}{T} = L^{-2}A$.
This corresponds to the unit $A/m^2$.