[Dirty]

Calculate the wavelength (in nm) of a photon emitted from a hydrogen atom when its electron drops from the n=3 state to the n=1 state.

I got 102 nm.
Ef - Ei=E =-13.6-(-1.5) = 12.1 eV
then used wave length=hc/E

1240eV x NM/12.1 eV = 102nm
Is this is correct way of doing it? or should I be using the formula 1/wave length=R(1/n^2f - 1/n^2i) ??

[Borek]

13.6-(-1.5) is not 12.1, it is 15.1.

No idea where did you get 13.6 and -1.5 values for energies - but if they were correct, results should be identical.

However somehow I don't think -1.5 is OK, negative energy?

[savy2020]

Ef - Ei = E = - 13.6 - (-1.5) = 12.1 eV

That should have been
E_f - E_i = - E = -13.6 - (1.5) = -12.1

Is this is correct way of doing it? or should I be using the formula 1/wave length=R(1/n^2f - 1/n^2i) ??

Both the ways you mentioned are correct.
Either way you get the same answer.

@Borek
Energy of an electron is negative in an atom since conventionally energy is considered "zero" when electron is at infinity. So in any orbit around the nucleus energy will be negative.
In the first orbit of H acc. to Bohrs's theory energy is calculated as -13.6eV.
and 1.5 ~ 13.6/9 as energy inversely varies as square of 'n'

[Borek]

Energy of an electron is negative in an atom since conventionally energy is considered "zero" when electron is at infinity. So in any orbit around the nucleus energy will be negative.
In the first orbit of H acc. to Bohrs's theory energy is calculated as -13.6eV.
and 1.5 ~ 13.6/9 as energy inversely varies as square of 'n'

That's just the matter of convention. However, in the original post (before it was edited and minus was added before 13.6 making me look like an idiot) energies differed in sign - and that was an obvious sign (pun not intended) that something is wrong.