Let the AP be $a_n=a+(n-1)d$ with first term $a$ and common difference $d$.

We are given two conditions:

1. $\sum_{n=1}^{20} \frac{1}{a_na_{n+1}}=\frac49$
2. Sum of $21$ terms is $189$.

1. Use the telescoping form of the given sum

For an AP, $a_{n+1}-a_n=d$ so $\frac{1}{a_na_{n+1}}=\frac{1}{d}\left(\frac{1}{a_n}-\frac{1}{a_{n+1}}\right)$ because $\frac{1}{d}\left(\frac{1}{a_n}-\frac{1}{a_{n+1}}\right)=\frac{1}{d}\cdot\frac{a_{n+1}-a_n}{a_na_{n+1}}=\frac{1}{a_na_{n+1}}.$

Hence, $\sum_{n=1}^{20}\frac{1}{a_na_{n+1}}=\frac{1}{d}\sum_{n=1}^{20}\left(\frac{1}{a_n}-\frac{1}{a_{n+1}}\right)$ which telescopes to $\frac{1}{d}\left(\frac{1}{a_1}-\frac{1}{a_{21}}\right)=\frac49.$

Now $a_1=a$ and $a_{21}=a+20d$, so $\frac{1}{d}\left(\frac{1}{a}-\frac{1}{a+20d}\right)=\frac49.$

Simplify: $\frac{1}{d}\cdot\frac{20d}{a(a+20d)}=\frac49$ $\frac{20}{a(a+20d)}=\frac49$ $a(a+20d)=45. \qquad (1)$

2. Use the sum of 21 terms

Sum of first $21$ terms of an AP is $S_{21}=\frac{21}{2}[2a+20d]=21(a+10d).$ Given $S_{21}=189$, $21(a+10d)=189$ $a+10d=9. \qquad (2)$

3. Find $a_6a_{16}$

Now, $a_6=a+5d,$ $a_{16}=a+15d.$

Their product is $a_6a_{16}=(a+5d)(a+15d).$ Rewrite around the middle term $a+10d$: $a+5d=(a+10d)-5d,$ $a+15d=(a+10d)+5d.$

So, $a_6a_{16}=[(a+10d)-5d][(a+10d)+5d]=(a+10d)^2-25d^2.$

But a better direct relation comes from $a(a+20d)=(a+10d-10d)(a+10d+10d)=(a+10d)^2-(10d)^2.$

Using (1) and (2): $45=9^2-100d^2$ $45=81-100d^2$ $100d^2=36$ $d^2=\frac{9}{25}.$

Therefore, $a_6a_{16}=(a+10d)^2-25d^2=9^2-25\cdot\frac{9}{25}=81-9=72.$

4. Check options

Thus, $a_6a_{16}=72.$ So the correct option is B.

5. Comparison with stored answer

Stored correct answer: B

Our derived answer: B

They agree.