The Correct Option is B
Solution and Explanation
Approach: Replace the scary $10^{50}+10^{25}$ with a small twin that behaves identically, find a digit-sum pattern, then scale up. Use $10^{6}+10^{3}-123$ as a model (same "$1$, gap of zeros, $1$, gap of zeros" shape).
Step 1: Small model. $10^{6}+10^{3} = 1001000.$ Subtract $123$: $1001000-123 = 1000877.$ Digit sum $= 1+0+0+0+8+7+7 = 23.$
Here the block of zeros between the two $1$s had length $2$, and below the lower $1$ there were $3$ zeros that turned into $877$ plus $0$ nines.
Step 2: See the structure. In general $10^{a}+10^{b}-123$ (with $a>b\ge 3$) becomes: a leading $1$, then $(a-b-1)$ zeros, then a $0$, then $(b-3)$ nines, then $877$.
Step 3: Apply with $a=50,\,b=25$.
Nines: $b-3 = 22$, contributing $22\times 9 = 198.$
Leading $1$ contributes $1.$
Tail $877$ contributes $8+7+7 = 22.$
All the in-between zeros contribute $0.$
Total $= 1+198+22 = \mathbf{221}.$ (Sanity check: the model gave $1+0+22 = 23$, matching $1000877$.)
\[ \text{Digit sum} = 221 \]