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C12H22O11 + H2O --> 4C2H5OH + 4CO2
No of moles of C12H22O11=GIVEN MASS/MOLAR MASS
                                            =3.42/342 = 0.01
According to the reaction,
1 mol of C12H22O11 requires 1 mol of H2O
But we have only 0.01 moles of C12H22O11. Hence, C12H22O11 is a limiting reagent.
So, C2H5OH will be formed only from amount of available sucrose.
From the reaction,
0.01 mol of C12H22O11 forms 4 moles of O   (solution formed-C2H50H)
==> 1mol of C12H22O11 forms=0.01*4=0.04mol of O
Since 1 mol of O contains 6.022*10^23 atoms
So, 0.04 mol of O will have=6.022*10^23 * 0.04=2.4 * 10^22 atoms of O.

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