The stepwise formation of $$[Cu(NH_3)_4]^{2+}$$ is given below:
$$Cu^{2+} + NH_3 \xrightleftharpoons{K_1} [Cu(NH_3)]^{2+}$$ 
$$[Cu(NH_3)]^{2+} + NH_3 \xrightleftharpoons{K_2} [Cu(NH_3)_2]^{2+}$$
$$[Cu(NH_3)_2]^{2+} + NH_3 \xrightleftharpoons{K_3} [Cu(NH_3)_3]^{2+}$$
$$[Cu(NH_3)_3]^{2+} + NH_3 \xrightleftharpoons{K_4} [Cu(NH_3)_4]^{2+}$$
The value of stability constants $$K_1$$, $$K_2$$, $$K_3$$ and $$K_4$$ are $$10^4$$, $$1.58 \times 10^3$$, $$5 \times 10^2$$ and $$10^2$$ respectively. The overall equilibrium constant for dissociation of $$[Cu(NH_3)_4]^{2+}$$ is $$x \times 10^{-12}$$. The value of $$x$$ is ______ (Rounded off to the nearest integer)

We are given the stepwise formation constants for $$[Cu(NH_3)_4]^{2+}$$: $$K_1 = 10^4$$, $$K_2 = 1.58 \times 10^3$$, $$K_3 = 5 \times 10^2$$, and $$K_4 = 10^2$$.

The overall stability constant for the formation of $$[Cu(NH_3)_4]^{2+}$$ is the product of all stepwise constants:

$$\beta_4 = K_1 \times K_2 \times K_3 \times K_4$$

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

Now we compute step by step:

$$= (1 \times 1.58 \times 5 \times 1) \times 10^{4+3+2+2}$$

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

The overall equilibrium constant for dissociation is the reciprocal of the formation constant:

$$K_{\text{dissociation}} = \frac{1}{\beta_4} = \frac{1}{7.9 \times 10^{11}}$$

$$= \frac{1}{7.9} \times 10^{-11} = 0.1266 \times 10^{-11}$$

$$= 1.266 \times 10^{-12} \approx 1 \times 10^{-12}$$

We are told the dissociation constant is $$x \times 10^{-12}$$. So:

$$x = 1$$

So, the answer is $$1$$.