Reaction of Fe2+ and S2- to form FeS
This is very poorly worded question, and really not just an equlibrium question. This is actually a solubility product question. Consider the given reaction:

$$\ce{Fe^{2+} (aq) + S^{2-} (aq) &lt;=&gt; FeS (s)} \tag{1}$$

If equal volumes of $\pu{0.06 M}$ of $\ce{Fe^2+}$ and $\pu{0.2 M}$ of $\ce{S^2-}$ are mixed, since the equilibrium constant for the reaction is $1.6 \times 10^{17}$ at $\pu{298K}$, we can assume that all $\ce{Fe^2+}$ are reacted because it is the limiting reagent and would give $\pu{0.06 M}\times V$ amount of solid $\ce{FeS}$ (where $V$ is the volume of $\ce{Fe^2+}$).

However, the author of the question ignore that $\ce{FeS}$ would gives solubility product of $\ce{Fe^2+}$ and $\ce{S^2-}$.

Since most sulfides of transition metals are insoluble in water, following equilibrium would establish after the original reaction:

$$\ce{FeS (s)  &lt;=&gt; F^2+ (aq) + S^2- (aq)}\text{;} \,\,\, K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]=4.9 \times 10^{-18} \tag{2}$$

Here, $K_\mathrm{sp}$ has not be provided, so I found it in here at $\pu{298K}$ ($\pu{25 ^\circ C}$). The concentration of $\ce{Fe^2+}$ at this equilibrium is given as $A \times 10^{-17}$. We can also calculate the concentration of $\ce{S^2-}$ from equation $(1)$ that $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ (since equal volumes has made final volume dobuled and hence concentration of remaining ions are halves). Since this value is much larger than the $[\ce{S^2-}]$ generated by dissociation of $\ce{FeS}$ ($\approx \sqrt{4.9 \times 10^{-18}}$), we can assume $\pu{0.07 M$ is the same concentration of $\ce{S^2-}$ at this equilibrium. Accordingly:

$$K_\mathrm{sp}= [\ce{F^2+}][\ce{S^2-}]= A \times 10^{-17} \times 0.07 = 4.9 \times 10^{-18}$$

$$\therefore \; A = \frac{4.9 \times 10^{-18}}{1.0 \times 10^{-17} \times 0.07}  = 7.0 $$

Note that this final answer is highly depends on the $K_\mathrm{sp}$ value. Thus, the deviation of this answer from your given answer is based on what value should have been given at the first place.

Suppose we ignore the solubility product as author may be suggesting: Then, once the original reaction is completed, significant amount of $\ce{S^2-}$ would remain unreacted since it is in excess (compared to $[\ce{Fe^2+}]$). That amount would be $\frac{\pu{(0.2-0.06) M}}{2}=\pu{0.07 M}$ as indicated above. Since at this point, $[\ce{Fe^2+}]$ is $A \times 10^{-17}$, the equilibrium constant is set up as follows as OP has done:

$$K = \frac{1}{[\ce{Fe^2+}][\ce{S^2-}]}=\frac{1}{[A \times 10^{-17}][0.07]}= 1.6 \times 10^{17}$$

$$\therefore \; A = \frac{1}{1.0 \times 0.07 \times 10^{-17}\times 1.6 \times 10^{17}}= 8.93$$

Although, this setup gives the expected answer, it is not the correct way to solve this (at least to my knowledge).