Calculating the value of equilibrium constant in terms of pressure
The problem lies in that our information of the system is in terms of concentration, so you actually calculated $K_c$.
$$\ce{2H2 +S2&lt;=&gt;2H2S}$$
$$\begin{array}{|l|l|l|}
\ce{H2 &amp; S2 &amp; H2S} \\ \hline
0.992\ \mathrm{M}&amp; 0\ \mathrm{M}&amp; 0.062\ \mathrm{M} \\
2(1.6\times 10^{-5}) &amp; (1.6\times 10^{-5}) &amp; -2(1.6\times 10^{-5}) \\
0.992+2(1.6\times 10^{-5})  &amp;  1.6\times10^{-5}\ \mathrm{M}   &amp;  0.062-2(1.6\times 10^{-5})\end{array}$$
$$K_c=\frac{(0.062\ \mathrm{M}-3.2\times 10^{-5}\ \mathrm{M})^2}{(1.6\times10^{-5}\ \mathrm{M})\cdot(0.992\ \mathrm{M}+3.2\times 10^{-5}\ \mathrm{M})^2}=245\ \mathrm{M^{-1}}$$
which is numerically the same as the value you obtained, except that I used a more precise molecular mass for $\ce{H2}$.
Using the relation $$K_p=K_c(RT)^{\Delta n},$$
where $$\Delta n=(\sum \text{coefficient of gaseous products})-(\sum\text{coefficient of gaseous reactants}),$$ we can calculate $K_p$:
$$K_p=(245\ \mathrm{M^{-1}})\cdot(8.3144598\ \mathrm{J\ K^{-1}\ mol^{-1}}\cdot 1670\ \mathrm{K})^{-1}=0.0176\ \mathrm{kPa^{-1}}$$