The units of ${\mathrm{K}}_{\mathrm{c}}$ for decomposition of ammonium bisulphide is

The molecular formula of ammonium bisulphide is  ${\mathrm{NH}}_4\mathrm{HS}$

The decomposition of the solid ammonium bisulphide is  $\left({\mathrm{NH}}_4\right)\mathrm{HS}\left(\mathrm{s}\right)\leftrightarrows {\mathrm{NH}}_3\left(\mathrm{g}\right)+{\mathrm{H}}_2\mathrm{S}\left(\mathrm{g}\right)$

So the equilibrium constant of this decomposition (with respect to concentration) is ${\mathrm{K}}_{\mathrm{c}}=\frac{\left[{\mathrm{NH}}_3\right]\left[{\mathrm{H}}_2\mathrm{S}\right]}{\left[{\mathrm{NH}}_4\mathrm{HS}\right]}......\left(1\right)$ 

$\left[{\mathrm{NH}}_4\mathrm{HS}\right]$ is taken $1$as it is in solid state.

So, the equation$\left(1\right)$ become ${\mathrm{K}}_{\mathrm{c}}=\left[{\mathrm{NH}}_3\right]\left[{\mathrm{H}}_2\mathrm{S}\right]$

Now the unit of concentration of ${\mathrm{NH}}_3$is $\left[{\mathrm{NH}}_3\right]=\mathrm{mol}.{\mathrm{L}}^{-1}$and the unit of concentration of ${\mathrm{H}}_2\mathrm{S}$ is $\left[{\mathrm{H}}_2\mathrm{S}\right]=\mathrm{mol}.{\mathrm{L}}^{-1}$ as for both the gases stoichiometric value is $1$.

So, the unit of ${\mathrm{K}}_{\mathrm{c}}$ become $\mathrm{mol}.{\mathrm{L}}^{-1}\times \mathrm{mol}.{\mathrm{L}}^{-1}={\mathrm{mol}}^2{\mathrm{L}}^{-2}$

Hence, option $2$ is the correct option.