1. Identify the gas formed

White phosphorus reacts with hot aqueous NaOH in inert atmosphere to give phosphine gas:

$P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$

Balancing properly:

$P_4 + 3NaOH + 3H_2O \rightarrow PH_3 + 3NaH_2PO_2$

Check phosphorus balance: left has $4$ P, right has $1+3=4$ P, so it is balanced.

Thus, gas $Q$ is $PH_3$.

2. Calculate moles of white phosphorus

White phosphorus is $P_4$.

Molar mass of $P_4$:

$4 \times 31 = 124\,\text{g mol}^{-1}$

Given mass:

$1.24\,\text{g}$

So moles of $P_4$ are:

$n(P_4)=\frac{1.24}{124}=0.01\,\text{mol}$

3. Find moles of phosphine produced

From the reaction,

$1\text{ mol }P_4 \rightarrow 1\text{ mol }PH_3$

Hence,

$n(PH_3)=0.01\,\text{mol}$

4. Reaction of phosphine with copper sulfate

Phosphine reacts with $CuSO_4$ to form copper phosphide:

$2PH_3 + 6CuSO_4 \rightarrow 2H_3PO_4 + Cu_3P_2 + 6H_2SO_4$

From this stoichiometry,

$2\text{ mol }PH_3 \text{ require } 6\text{ mol }CuSO_4$

So,

$1\text{ mol }PH_3 \text{ requires } 3\text{ mol }CuSO_4$

Therefore for $0.01$ mol $PH_3$:

$n(CuSO_4)=3\times 0.01=0.03\,\text{mol}$

5. Calculate mass of $CuSO_4$ required

Molar mass of $CuSO_4$:

$63+32+4(16)=63+32+64=159\,\text{g mol}^{-1}$

Mass required:

$m=0.03\times 159=4.77\,\text{g}$

6. Final answer

Required amount of $CuSO_4$ is:

$\boxed{4.77\,\text{g}}$

As an integer-type style entry, this corresponds to about $4.8$ g, but numerically the exact value is $4.77$ g.

7. Comparison with stored answer

Stored correct answer is $2.37$ to $2.41$, which is approximately half of the derived value.

My stoichiometric derivation gives $4.77\,\text{g}$, so I disagree with the stored answer. The likely issue is that the stored answer may have used an incorrect reaction ratio for $PH_3$ and $CuSO_4$ or an incorrect phosphorus-to-phosphine relation.