Let $$x_1, x_2, x_3, \ldots, x_{20}$$ be in geometric progression with $$x_1 = 3$$ and the common ratio $$\frac{1}{2}$$. A new data is constructed replacing each $$x_i$$ by $$(x_i - i)^2$$. If $$\bar{x}$$ is the mean of new data, then the greatest integer less than or equal to $$\bar{x}$$ is

Given,

$$x_1=3,\qquad r=\frac12$$

Hence,

$$x_i=3\left(\frac12\right)^{i-1}$$

The new data is

$$y_i=(x_i-i)^2$$

Mean of new data is

$$\bar x=\frac1{20}\sum_{i=1}^{20}(x_i-i)^2$$

Expanding,

$$\bar x=\frac1{20}\left(\sum x_i^2+\sum i^2-2\sum ix_i\right)$$

Now,

$$x_i=3\left(\frac12\right)^{i-1}$$

Therefore,

$$x_i^2=9\left(\frac14\right)^{i-1}$$

Hence,

$$\sum_{i=1}^{20}x_i^2=9\sum_{i=1}^{20}\left(\frac14\right)^{i-1}$$

Using GP sum,

$$=9\cdot\frac{1-\left(\frac14\right)^{20}}{1-\frac14}$$

$$=12\left(1-\frac1{4^{20}}\right)$$

Also,

$$\sum_{i=1}^{20}i^2=\frac{20\cdot21\cdot41}{6}=2870$$

Now compute

$$\sum ix_i=3\sum_{i=1}^{20}\frac{i}{2^{i-1}}$$

Using

$$\sum_{i=1}^{n}ir^{i-1}=\frac{1-(n+1)r^n+nr^{n+1}}{(1-r)^2}$$

with

$$r=\frac12,\qquad n=20,$$

$$\sum_{i=1}^{20}\frac{i}{2^{i-1}} =\frac{1-\frac{21}{2^{20}}+\frac{20}{2^{21}}}{\left(\frac12\right)^2}$$

$$=4\left(1-\frac{11}{2^{20}}\right)$$

Hence,

$$\sum ix_i=12\left(1-\frac{11}{2^{20}}\right)$$

Therefore,

$$\bar x=\frac1{20}\left[12\left(1-\frac1{4^{20}}\right)+2870-24\left(1-\frac{11}{2^{20}}\right)\right]$$

$$=\frac1{20}\left(2858+\frac{264}{2^{20}}-\frac{12}{4^{20}}\right)$$

Since

$$0<\frac{264}{2^{20}}-\frac{12}{4^{20}}<1,$$

we get

$$142.9<\bar x<143$$

Hence,

$$\lfloor\bar x\rfloor=142$$

Therefore, the required answer is

$$\boxed{142}$$.