Dimensional formula for thermal conductivity is (here K denotes the temperature):

We begin with the physical definition of thermal conductivity, usually denoted by the symbol $$k$$ (some books use $$K$$). According to Fourier’s law of heat conduction, the rate at which heat flows through a slab is proportional to the area, the temperature gradient, and the thermal conductivity. In symbols, the law is written as

$$\frac{Q}{t}=k\,A\,\frac{\Delta T}{\Delta x}$$

where

$$Q$$ = total heat conducted (a form of energy),
$$t$$ = time taken for that conduction,
$$A$$ = cross-sectional area of the slab,
$$\Delta T$$ = temperature difference across the slab, and
$$\Delta x$$ = thickness of the slab.

Re-arranging this expression to isolate $$k$$, we get

$$k=\frac{Q\,\Delta x}{A\,\Delta T\,t}$$

To obtain the dimensional formula of $$k$$, we now replace every quantity by its basic dimensions.

First, we recall that heat $$Q$$ is a form of energy, and the dimensional formula for energy is the same as that for work: $$[Q]=ML^{2}T^{-2}$$.

The remaining quantities have the following dimensions:

$$[\Delta x]=L,\qquad [A]=L^{2},\qquad [\Delta T]=K,\qquad [t]=T.$$

Substituting each of these into the expression for $$k$$ gives

$$[k]=\frac{[Q]\,[\Delta x]}{[A]\,[\Delta T]\,[t]} =\frac{\left(ML^{2}T^{-2}\right)\,L}{L^{2}\,K\,T}$$

Now we simplify the powers of $$L$$ and $$T$$ step by step:

$$[k]=\frac{M\,L^{2}T^{-2}\,L}{L^{2}\,K\,T} =\frac{M\,L^{3}\,T^{-2}}{L^{2}\,K\,T}$$

Dividing the exponents of like dimensions in numerator and denominator, we obtain

$$[k]=M\,L^{3-2}\,T^{-2-1}\,K^{-1} =M\,L^{1}\,T^{-3}\,K^{-1}$$

This can be written compactly as

$$[k]=MLT^{-3}K^{-1}$$

Comparing this with the given options, we see that it matches Option D.

Hence, the correct answer is Option D.