Let the maximum concentration of each solution be $x$ $mol/L$. After mixing, the volume of the concentrations of each solution will be reduced to half i.e., $x / 2$.

$\therefore\left[ FeSO _{4}\right]=\left[ Na _{2} S \right]=\frac{x}{2} M$

Then, $\left[ Fe ^{2+}\right]=\left[ FeSO _{4}\right]=\frac{x}{2} M$

Also, $\left[ S ^{2-}\right]=\left[ Na _{2} S \right]=\frac{x}{2} M$

$Fe{S_{(s)}} \leftrightarrow Fe_{(aq)}^{2 + } + S_{\left( {aq} \right)}^{2 – }$

$K_{ sp }=\left[ Fe ^{2+}\right]\left[ S ^{2-}\right]$

$6.3 \times 10^{-18}=\left(\frac{x}{2}\right)\left(\frac{x}{2}\right)$

$\frac{x^{2}}{4}=6.3 \times 10^{-18}$

$\Rightarrow x=5.02 \times 10^{-9}$

If the concentrations of both solutions are equal to or less than $5.02 \times 10^{-9} \,M ,$ then there will be no precipitation of iron sulphide.