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

Then the $r^{\text{th}}$ term is $T_r=a+(r-1)d.$

We are given:

1. $T_m=\dfrac{1}{25}$
2. $T_{25}=\dfrac{1}{20}$
3. $20\sum_{r=1}^{25} T_r=13$

We need to find $5m\sum_{r=m}^{2m} T_r.$

1. Use the sum of first 25 terms

Since $20\sum_{r=1}^{25} T_r=13,$ we get $\sum_{r=1}^{25} T_r=\frac{13}{20}.$

Now, $S_{25}=\frac{25}{2}\left[2a+24d\right]=25(a+12d).$ So, $25(a+12d)=\frac{13}{20}$ $a+12d=\frac{13}{500}. \qquad (1)$

2. Use $T_{25}=\dfrac{1}{20}$

$T_{25}=a+24d=\frac{1}{20}. \qquad (2)$

Subtract (1) from (2): $(a+24d)-(a+12d)=\frac{1}{20}-\frac{13}{500}$ $12d=\frac{25-13}{500}=\frac{12}{500}$ $d=\frac{1}{500}.$

Substitute into (1): $a+12\cdot \frac{1}{500}=\frac{13}{500}$ $a=\frac{1}{500}.$

Hence, $T_r=a+(r-1)d=\frac{1}{500}+\frac{r-1}{500}=\frac{r}{500}.$

3. Use $T_m=\dfrac{1}{25}$

Since $T_m=\frac{m}{500}=\frac{1}{25},$ we get $m=20.$

4. Compute $\sum_{r=m}^{2m} T_r$

Here $m=20$, so we need \sum_{r=20}^{40} T_r=\sum_{r=20}^{40} \frac{r}{500}=rac{1}{500}\sum_{r=20}^{40} r.

Now, \sum_{r=20}^{40} r=\frac{(20+40)(40-20+1)}{2}=rac{60\cdot 21}{2}=630.

Therefore, $\sum_{r=20}^{40} T_r=\frac{630}{500}=\frac{63}{50}.$

5. Find the required value

$5m\sum_{r=m}^{2m} T_r=5\cdot 20\cdot \frac{63}{50}.$

Since $5\cdot 20=100$, $100\cdot \frac{63}{50}=2\cdot 63=126.$

6. Check options

The value is $\boxed{126}.$

So the correct option is B.