The dimension of electrical conductivity is ________.

Step 1: Relate electrical conductivity to current density and electric field.

Ohm's law in microscopic form relates current density $\vec{J}$, electrical conductivity $\sigma$, and electric field $\vec{E}$.

$$\vec{J} = \sigma \vec{E}$$ $$\sigma = \frac{J}{E}$$

Step 2: Express current density and electric field in terms of fundamental quantities.

Current density $J$ is current per unit area, and electric field $E$ is force per unit charge.

$$J = \frac{I}{A}$$ $$E = \frac{F}{q}$$

Step 3: Determine the dimensional formula of current density $[J]$.

Substitute the dimensions of current $[I] = I$ and area $[A] = L^{2}$.

$$[J] = \frac{[I]}{[A]}$$ $$[J] = \frac{I}{L^{2}}$$ $$[J] = I L^{-2}$$

Step 4: Determine the dimensional formula of electric field $[E]$.

Substitute the dimensions of force $[F] = M L T^{-2}$ and charge $[q] = I T$.

$$[E] = \frac{[F]}{[q]}$$ $$[E] = \frac{M L T^{-2}}{I T}$$ $$[E] = M L T^{-3} I^{-1}$$

Step 5: Calculate the dimensional formula of electrical conductivity $[\sigma]$.

Substitute the dimensions of $[J]$ and $[E]$ into the expression for $[\sigma]$.

$$[\sigma] = \frac{[J]}{[E]}$$ $$[\sigma] = \frac{I L^{-2}}{M L T^{-3} I^{-1}}$$ $$[\sigma] = M^{-1} L^{-3} T^{3} I^{2}$$

Answer: $M^{-1} L^{-3} T^{3} I^{2}$