$\left[ Cu \left( NH _{3}\right)\right]^{2+}+ NH _{3} \stackrel{ K ,}{\rightleftharpoons}\left[ Cu \left( NH _{3}\right)_{2}\right]^{2+}$

$\left[ Cu \left( NH _{3}\right)_{2}\right]^{2+}+ NH _{3} \stackrel{ K _{3}}{\rightleftharpoons}\left[ Cu \left( NH _{3}\right)_{3}\right]^{2+}$

$\left[ Cu \left( NH _{3}\right)_{3}\right]^{2+}+ NH _{3} \stackrel{ K _{ a }}{\rightleftharpoons}\left[ Cu \left( NH _{3}\right)_{4}\right]^{2+}$

$Cu ^{2+}+4 NH _{3} \stackrel{ K }{\rightleftharpoons}\left[ Cu \left( NH _{3}\right)_{4}\right]^{2+}$

So

$K = K _{1} \times K _{2} \times K _{3} \times K _{4}$

$=10^{4} \times 1.58 \times 10^{3} \times 5 \times 10^{2} \times 10^{2}$

$K =7.9 \times 10^{11}$

Where $K \rightarrow$ Equilibrium constant for formation of $\left[ Cu \left( NH _{3}\right)_{4}\right]^{2+}$

So equilibrium constant $\left( K ^{\prime}\right)$ for dissociation

of $\left[ Cu \left( NH _{3}\right)_{4}\right]^{2+}$ is $\frac{1}{ K }$

$K ^{\prime}=\frac{1}{ K }$

$K ^{\prime}=\frac{1}{7.9 \times 10^{11}}$

$=1.26 \times 10^{-12}=\left( x \times 10^{-12}\right)$

So the value of $x=1.26$

$OMR$ Ans $=1$ (After rounded off to the nearest integer)