In the following $$I$$ refers to current and other symbols have their usual meaning. Choose the option that corresponds to the dimensions of electrical conductivity:

We have to find the dimensional formula of electrical conductivity, usually denoted by $$\sigma$$. By definition electrical conductivity is the reciprocal of electrical resistivity $$\rho$$, so

$$\sigma=\dfrac{1}{\rho}.$$

Therefore, once we know the dimensions of $$\rho$$ we can simply invert them to obtain those of $$\sigma$$.

The resistivity $$\rho$$ is related to resistance $$R$$ through the well-known relation

$$\rho=\dfrac{RA}{l},$$

where $$A$$ is the cross-sectional area of the conductor and $$l$$ is its length. Next, resistance $$R$$ itself is connected to potential difference $$V$$ and current $$I$$ by Ohm’s law, which states first:

$$R=\dfrac{V}{I}.$$

So we must obtain the dimensions of $$V$$. Potential difference (voltage) is defined as work done per unit charge, i.e.

$$V=\dfrac{\text{Work}}{\text{Charge}}.$$

Work (or energy) has the mechanical dimensions $$ML^2T^{-2}$$, while charge is current $$I$$ multiplied by time $$T$$. Thus

$$[\!V\!]=\dfrac{ML^2T^{-2}}{IT}=ML^2T^{-3}I^{-1}.$$

Substituting this in Ohm’s law we get the dimensions of resistance:

$$[R]=\dfrac{[V]}{[I]}=\dfrac{ML^2T^{-3}I^{-1}}{I}=ML^2T^{-3}I^{-2}.$$

Now we insert this result in the expression for resistivity:

$$[\rho]=[R]\,[A]\,[l]^{-1}.$$

The area $$A$$ has the dimensions of length squared $$L^{2}$$, while length $$l$$ has dimensions $$L$$. Therefore,

$$[\rho]=\left(ML^2T^{-3}I^{-2}\right)\left(L^{2}\right)L^{-1}=ML^{3}T^{-3}I^{-2}.$$

Finally, conductivity is the reciprocal of resistivity, so we take the inverse of every exponent:

$$[\sigma]=[\rho]^{-1}=M^{-1}L^{-3}T^{3}I^{2}.$$

Looking back at the given options, this matches exactly with Option B.

Hence, the correct answer is Option B.