Consider the following reactions:
NaCl + K$$_2$$Cr$$_2$$O$$_7$$ + H$$_2$$SO$$_4$$ (Conc.) $$\rightarrow$$ (A) + side products
(A) + NaOH $$\rightarrow$$ (B) + side products
(B) + H$$_2$$SO$$_4$$ + H$$_2$$O$$_2$$ (dilute) $$\rightarrow$$ (C) + side products
The sum of the total number of atoms in one molecule each of (A), (B) and (C) is

We have first reaction involving common salt, sodium chloride, with potassium dichromate and concentrated sulphuric acid.

From qualitative inorganic analysis it is well-known that $$\text{Cl}^-$$ in the presence of $$\text{K}_2\text{Cr}_2\text{O}_7$$ and conc. $$\text{H}_2\text{SO}_4$$ produces deep-red vapours of chromyl chloride. Writing the balanced equation,

$$4\,\text{NaCl}+\,\text{K}_2\text{Cr}_2\text{O}_7+3\,\text{H}_2\text{SO}_4 \;\longrightarrow\;2\,\text{CrO}_2\text{Cl}_2+\,\text{K}_2\text{SO}_4+2\,\text{Na}_2\text{SO}_4+3\,\text{H}_2\text{O}$$

The substance asked for is the red vapour $$\text{CrO}_2\text{Cl}_2$$, so

$$(A)=\text{CrO}_2\text{Cl}_2\;(\text{chromyl chloride}).$$

Counting the atoms present in one molecule of (A):

$$\text{Cr}=1,\;\text{O}=2,\;\text{Cl}=2\;\Longrightarrow\;1+2+2=5\text{ atoms.}$$

Now (A) is treated with aqueous sodium hydroxide. The standard reaction is

$$\text{CrO}_2\text{Cl}_2+4\,\text{NaOH}\;\longrightarrow\;\text{Na}_2\text{CrO}_4+2\,\text{NaCl}+2\,\text{H}_2\text{O}.$$

The yellow compound obtained is sodium chromate, therefore

$$(B)=\text{Na}_2\text{CrO}_4\;(\text{sodium chromate}).$$

Next, (B) is treated with dilute $$\text{H}_2\text{SO}_4$$ and hydrogen peroxide. In acidic medium the chromate ion $$\text{CrO}_4^{2-}$$ reacts with $$\text{H}_2\text{O}_2$$ to give the deep-blue peroxo compound generally written as $$\text{CrO}_5$$. A simplified ionic equation is

$$\text{CrO}_4^{2-}+4\,\text{H}^++\text{H}_2\text{O}_2\;\longrightarrow\;\text{CrO}_5+3\,\text{H}_2\text{O}.$$

Because the sodium ions become spectator ions after acidification, the substance actually isolated is

$$(C)=\text{CrO}_5\;(\text{peroxochromic acid, transient}).$$

Counting atoms in one molecule of (B):

$$\text{Na}=2,\;\text{Cr}=1,\;\text{O}=4\;\Longrightarrow\;2+1+4=7\text{ atoms.}$$

Counting atoms in one molecule of (C):

$$\text{Cr}=1,\;\text{O}=5\;\Longrightarrow\;1+5=6\text{ atoms.}$$

Finally, summing the total number of atoms contained in one molecule each of (A), (B) and (C):

$$5\;(\text{from }A)+7\;(\text{from }B)+6\;(\text{from }C)=18.$$

So, the answer is $$18$$.