Above answer was was given accidently and I can't figure out how to delete it so here is the correct answer:
In the given reaction Oxidation numbers of the elements are:
LHS:
HNO3==H(+1)N(+5)O(-2)
H2S==H(+1)S(-2)
RHS:
H2O==H(+1)O(-2)
NO==N(+2)O(-2)
S==S(0)
It is clear from observing the oxidation numbers that H2S is oxidised and HNO3 is reduced
By balancing using the ion electron method the 2 half reactions we get are:
3H2S--->3S+6e-+6H+
6H++6e-+2HNO3--->2NO+4H2O
We can see that 6 electrons have been exchanged in the reaction therefore n-factor=6
Equivalent weight=Molar mass/n-factor
=126/6=21