c
According to Faraday's first law of electrolysis

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

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 \((+3)\)