Unit derivation concerning Nernst equation We know that Nernst equation is given by:

$E=E^{\circ}+\frac{RT}{nF} \ln[\ce{M^{n+}}]$.

Now, A question asks me to derive the unit of  $\frac{RT}{nF}$.

Which I think can easily be derived this way:

$\frac{RT}{nF}=\frac{E-E^{\circ}}{\ln[\ce{M^{n+}}]}$,

Now, the unit of $E-E^{\circ}$ is volts (since it is emf), and $\ln[\ce{M^{n+}}]$ is just a constant numerical value.

Hence, I think the unit of $\frac{RT}{nF}$ is Volts.

This was my approach. Can anyone tell me if it is a correct approach?
M. Farooq have given good description of how to analyse units in a equation. Therefore, your assumption of the unit of $\dfrac{RT}{nF}$ should be Volts is correct. However, your question is not to assume but to derive. What's that means is unit conversion. Let's see the units of all four variables:

$R$ is a constant: $\pu{8.314 J mol-1 K-1}$
$F$ is also a constant: $\approx \pu{96485  C mol-1}$
$T$ is a variable: $\pu{K}$
$n$ is number of total electrons in the redox reaction involved, and hence a variable but unit less.

Now lets see the unit conversion:
$$\frac{RT}{nF} = \frac{\pu{J mol-1 K-1}\times \pu{K}}{\pu{C mol-1}} = \frac{\pu{J}}{\pu{C}} $$
By definition (Ohm's law):
$$ V = \frac{\text{potential energy}}{\text{Charge}} = \frac{\pu{J}}{\pu{C}} $$
$$\therefore \:\; \frac{RT}{nF} = \frac{\pu{J}}{\pu{C}}= \pu{V} $$