Total number of sulphate ions present in $3.92 g$ of chromic sulphate is $(Cr=52, S=32,O=16)$

Chemical formula of chromic sulphate i.e ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$ .

Here one mole chromic sulphate contains 3 moles of sulphate ions.

Molecular mass of  ${\mathrm{Cr}}_2{\left({\mathrm{SO}}_4\right)}_3$ = 392g

No of moles of chromic sulphate= $3.92/392$=0.01 mole

Now using unitary method, 0.01 mole chromic sulphate contains 0.03 mole sulphate ions

Now one mole= in ${\mathrm{N}}_0=6.022$ $\times {10}^{23}$ i.e., Avogadro's number entities

Thus no of sulphate ions in 0.03mole sulphate ions=$0.03\times N$ = $1.8\times {10}^{22}$

Hence option (A) is the correct answer.