Let the A.P. be $a,\, a+d,\, a+2d,\, \dots$ where $a$ is the first term and $d$ is the common difference.

Use the sum of the first three terms Given: $a+(a+d)+(a+2d)=54$ $3a+3d=54$ $a+d=18$

But the second term of the A.P. is $a+d$, so the second term is $18$.

Use the condition on the sum of the first 20 terms Sum of first $20$ terms: $S_{20}=\frac{20}{2}[2a+19d]=10(2a+19d)$

Since $a+d=18$, we have: $a=18-d$

Substitute into $S_{20}$: $S_{20}=10\big(2(18-d)+19d\big)$ $S_{20}=10(36-2d+19d)$ $S_{20}=10(36+17d)=360+170d$

Given: $1600<S_{20}<1800$ so, $1600<360+170d<1800$

Solve the inequality Subtract $360$ throughout: $1240<170d<1440$

Divide by $170$: $\frac{1240}{170}<d<\frac{1440}{170}$ $7.294\dots<d<8.470\dots$

Since the A.P. consists of positive integers, $d$ must be an integer. Hence, $d=8$

Find the first term $a=18-d=18-8=10$

Find the 11th term $T_{11}=a+10d=10+10(8)=90$

Check with options $T_{11}=90$ So the correct option is B.