$W\,=\,\frac {E\times i\times t}{96500}$

Where $E\,=$ equivqlent weight

$=\,\frac {mol.\,mass\,of\,metal\,(M)}{oxidation\,state\,of\,metal(x)}$

Substituting the value in  the formula

$W\,=\,\frac {M}{x}$  $\times $ $\frac {i\times t}{96500}$

or $x=\frac {M}{W}$  $\times $ $\frac {i\times t}{96500}$ $ = \,\frac{{10\, \times \,2 \times 60 \times 60\,}}{{96500\, \times 0.250}} = 3$

[Given :no . of moles $=\frac {M}{W}\,=\,0.250$ ]

Hence oxidation state of metals is $(+3)$