1. Use the relation between conductivity and resistivity

Electrical conductivity $\sigma$ is the reciprocal of resistivity $\rho$:

$\sigma = \frac{1}{\rho}$

So we first find the dimensions of resistivity.

2. Dimensions of resistance

From Ohm’s law,

$V = IR \Rightarrow R = \frac{V}{I}$

Now,

$V = \frac{\text{work}}{\text{charge}}$

Dimensions of work:

$[W] = ML^2T^{-2}$

Dimensions of charge:

$[Q] = IT$

Hence dimensions of potential difference:

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

Therefore,

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

3. Dimensions of resistivity

Using

$R = \rho \frac{l}{A} \Rightarrow \rho = R\frac{A}{l}$

where area $A$ has dimensions $L^2$ and length $l$ has dimensions $L$.

So,

$[\rho] = [R]\cdot \frac{L^2}{L} = [R]\cdot L$

$[\rho] = ML^2T^{-3}I^{-2} \cdot L = ML^3T^{-3}I^{-2}$

4. Dimensions of conductivity

Since

$\sigma = \frac{1}{\rho}$

we get

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

5. Match with the options

The required dimensions are:

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

This matches Option C.

6. Verification with stored answer

Stored correct answer: C

Derived answer: C

So the derived answer agrees with the stored answer.