1. Reaction of $\mathrm{Cu(NO_3)_2}$ with excess KI

Copper(II) ions oxidize iodide ions to iodine, and themselves get reduced to copper(I) iodide precipitate.

The reaction is:

$2\,\mathrm{Cu^{2+}} + 4\,\mathrm{I^-} \rightarrow 2\,\mathrm{CuI}(s) + \mathrm{I_2}$

So, from $2$ mol of $\mathrm{Cu^{2+}}$, we get $1$ mol of $\mathrm{I_2}$.

The brown solution is due to dissolved $\mathrm{I_2}$.

2. Moles of $\mathrm{Cu(NO_3)_2}$

Molar mass of $\mathrm{Cu(NO_3)_2}$:

$= 63 + 2(14 + 3\times 16)$ $= 63 + 2(14+48)$ $= 63 + 124 = 187$

Thus,

$n\big(\mathrm{Cu(NO_3)_2}\big)=\frac{3.74}{187}=0.02\text{ mol}$

So, moles of $\mathrm{Cu^{2+}} = 0.02$ mol.

3. Moles of iodine formed

Using

$2\,\mathrm{Cu^{2+}} \rightarrow \mathrm{I_2}$

Therefore,

$n(\mathrm{I_2})=\frac{0.02}{2}=0.01\text{ mol}$

4. Reaction of iodine with $\mathrm{H_2S}$

Hydrogen sulfide reduces iodine to iodide and itself gets oxidized to sulfur:

$\mathrm{I_2 + H_2S \rightarrow 2HI + S}(s)$

Thus, $1$ mol of $\mathrm{I_2}$ gives $1$ mol of sulfur precipitate $X$.

So,

$n(X)=n(S)=0.01\text{ mol}$

5. Mass of precipitate $X$

Since $X = \mathrm{S}$,

$m = n\times M = 0.01 \times 32 = 0.32\text{ g}$

6. Final Answer

$\boxed{0.32\text{ g}}$

As an integer-type entry style, the numerical value is:

$\boxed{32}\text{ (if entered as }0.32\text{ g, value is }0.32)$

Here the asked amount in grams is clearly $\boxed{0.32\text{ g}}$.