The correct option is C 16.5 Ω
Equivalent resistance for series connection is:
Req=R1+R2+R3+...+Rn
Since 12 Ω & 18 Ω are in series, we can can calculate the equivalent resistance for this combination, R1eq=12+18=30Ω
For parallel connection, equivalent resistance,
1Req=1R1+1R2+1R3+...+1Rn
Now, R1eq & 10 Ω are in parallel
Therefore, equivalent resistance for this combination is:
∴1R2eq=1R1eq+110=130+110=(1+330)
⇒1R2eq=430
⇒R2eq=304=7.5 Ω
Now, 9 Ω & R2eq are in series.
∴ Equivalent resistance for this combination is:
R3eq=9+R2eq=9+7.5=16.5 Ω.