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In the one mole of $Al_2(CO_3)_3$ there are $N_A$ formula units, i.e. $6.022xx10^23$ formula units.

$"Moles of aluminum carbonate"$ $=$ $"Mass"/"Molar mass"$

$=$ $(47.6*g)/(233.99*g*mol^-1)$ $=$ $0.203*mol$

But in each mole of the carbonate there are clearly $9$ $mol$ $O$ atoms.

And thus there are $9xx0.203*mol$ $O$ atoms in the given quantity.

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