In the following reaction :

$3Fe+4H_{2}O\to {\mathrm{Fe}}_3{\mathrm{O}}_4+4{\mathrm{H}}_2$, if the atomic weight of iron is 56, then iron equivalent weight in Fe₃O₄ its equivalent weight will be

$F{e}_3{O}_4$ is an Oxide. Oxidation state of Oxygen in Oxides is -2. So, total Oxidation state of four Oxygens is -8.

It implies that total Oxidation state of three Irons in $F{e}_3{O}_4$is +8. 

3 Iron atoms from zero O.S. got changed to +8 by the loss of 8 electrons.

$Equivalent weight of Iron in F{e}_3{O}_4=\frac{Mass of all Iron atoms 56\times 3}{Total number of e^{-1} lost by them 8}$=21