Let the first term of the A.P. be $a$ and common difference be $d$.

The $n$th term of an A.P. is $T_n=a+(n-1)d$

Given: $T_{10}=a+9d=\frac{1}{20}$ $T_{20}=a+19d=\frac{1}{10}$

Subtract the first equation from the second: $10d=\frac{1}{10}-\frac{1}{20}=\frac{1}{20}$ $d=\frac{1}{200}$

Substitute into $a+9d=\frac{1}{20}$: $a+9\left(\frac{1}{200}\right)=\frac{1}{20}$ $a+\frac{9}{200}=\frac{10}{200}$ $a=\frac{1}{200}$

Sum of first $n$ terms of an A.P.: $S_n=\frac{n}{2}[2a+(n-1)d]$

For $n=200$: $S_{200}=\frac{200}{2}[2\cdot \frac{1}{200}+199\cdot \frac{1}{200}]$ $=100\left(\frac{2+199}{200}\right)$ $=100\cdot \frac{201}{200}$ $=\frac{201}{2}=100\frac{1}{2}$

Therefore, the correct option is: $\boxed{B}$