1. We need the dimensions of $\left(\varepsilon_0\frac{d\Phi_E}{dt}\right).$

2. Use Gauss's law for electric flux through a closed surface: $\Phi_E = \oint \vec E\cdot d\vec A = \frac{q_{\text{encl}}}{\varepsilon_0}.$

So, $\varepsilon_0 \Phi_E = q.$

3. Differentiate with respect to time: $\varepsilon_0 \frac{d\Phi_E}{dt} = \frac{dq}{dt}.$

But $\frac{dq}{dt} = I,$ where $I$ is electric current.

4. Therefore, the dimensions of $\left(\varepsilon_0\frac{d\Phi_E}{dt}\right)$ are the same as those of electric current.

5. Checking options:

• A: electric charge $\to$ incorrect
• B: electric field $\to$ incorrect
• C: electric current $\to$ correct
• D: electric potential $\to$ incorrect

Hence, the correct option is C.