Step 1: Look for a 9 H branch
$\frac{18}{11}=\frac{2\times9}{2+9}$, so we need 2 H in parallel with 9 H, where $9=3+6$.
Step 2: Match the figure
Only (S) has $L_1$ alone on one path and $L_2$ with $L_3$ in series on the other path. So (S), option (B).
Final Answer:
Figure (S) gives 18/11 H, option (B).
\[ \boxed{S} \]