The sum of all the digits of the number $$(10^{50}+10^{25}-123)$$, is

$$(10^{50} + 10^{25} - 123)$$ can be written as $$10^{50} + (10^{25}-123)$$

$$1000-123$$ gives $$877$$. 
$$10000-123$$ gives $$9877$$.
$$100000-923$$ gives $$99877$$.
$$1000000-923$$ gives $$999877$$.
And so on.

Considering $$10^n-123$$, which has a total of $$n$$ digits; in general, of the $$n$$ total digits; $$2$$ are the digit $$7$$, $$1$$ is the digit $$8$$, and $$(n-3)$$ are the digit $$9$$.

Also, when added to $$10^{50}$$, the cumulative increase in the sum of the digit of the result would be by $$1$$, this would come from the leftmost digit of the result. All newly added digits will be zero.

Thus, the sum of the digits of the result will be given by;

$$[7*2] + [1*8] + [(25-3)*9] + [25*0] + [1*1] = 14+8+198+1 = 221$$

The correct answer is option B.