The quantity of electricity needed to electrolyse completely 1 M solution of ${\mathrm{CuSO}}_4,{\mathrm{Bi}}_2{\left({\mathrm{SO}}_4\right)}_3,{\mathrm{AlCl}}_3$ and ${\mathrm{AgNO}}_3$ each will be 

${\mathrm{CuSO}}_4+2{\mathrm{e}}^{-}\to \mathrm{Cu}+{\mathrm{SO}}_4^{-2}$

${\mathrm{Bi}}_2{\left({\mathrm{SO}}_4\right)}_3+6{\mathrm{e}}^{-}\to 2\mathrm{Bi}+3{\mathrm{SO}}_4^{-2}$

${\mathrm{AlCl}}_3+3{\mathrm{e}}^{-}\to \mathrm{Al}+3{\mathrm{Cl}}^{-}$

${\mathrm{AgNO}}_3+{\mathrm{e}}^{-}\to \mathrm{Ag}+{\mathrm{NO}}_3^{-}$

Since, in 1 M ${\mathrm{CuSO}}_4$ solution, 1 M ${\mathrm{Bi}}_2{\left({\mathrm{SO}}_4\right)}_3$ solution, 1 M ${\mathrm{AlCl}}_3$ solution and 1 M ${\mathrm{AgNO}}_3$ solution, 2 moles electron, 6 moles electron, 3 moles electron are needed to deposit Cu, Bi, Al and Ag at the cathode respectively. 

But one mole electron = 1 F electricity 

That's why number of faradays required to deposit 1 M of each $CuS{O}_4,B{i}_2{\left(S{O}_4\right)}_3,AlC{l}_3$ and $AgN{O}_3$ solution are 2 F, 6 F, 3 F and 1 F respectively 

Alternatively

Number of moles = Number of equivalents $\times$Valency

1 equivalent is deposited by 1 Faraday