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 $$9877$$.
$$100000-123$$ gives $$99877$$.
$$1000000-123$$ gives $$999877$$.
And so on.

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

Also, when added to $$10^{50}$$, the cumulative increase in the sum of the digits of the resultant number would be $$1$$, which would come from the leftmost digit of the result. All other 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.