The dimensions of the displacement current _0 (d _E)/(dt) are...

The displacement current is given by the expression $I_d = \varepsilon_0 \dfrac{d\Phi_E}{dt}$.

By definition, displacement current is a type of current, so its dimensions must be identical to those of an electric current.

We can also verify this dimensionally:

Electric flux $\Phi_E = E \times A = \dfrac{F}{q} \times A$

Permittivity of free space $\varepsilon_0 = \dfrac{q^2}{4\pi F r^2}$

Dimensions of $\varepsilon_0 \Phi_E = \left[ \dfrac{q^2}{F L^2} \right] \left[ \dfrac{F L^2}{q} \right] = [q]$

Dimensions of $\varepsilon_0 \dfrac{d\Phi_E}{dt} = \left[ \dfrac{q}{t} \right] = [A]$

Thus, the dimensions are identical to those of an electric current.