Assuming all bulbs are identical, rank the brightness of the bulbs, from brightest to dimmest.A)...

The right answer is (D) A > D=H > B=C > E=F=G
Brightness of the bulb $\propto$ current through the bulb
Let R represent the resistance of each bulb
Regarding the branch containing bulbs A, B, and C
$$\begin{gathered} R_{eq}=R+[\frac{R\times R}{R+R}] \\ =R+\frac{R}{2}=\frac{3R}{2} \\ i=\frac{2E}{3R} \end{gathered}$$
Current through A$=\frac{2E}{3R}$
Current through B and C A$$\begin{gathered} =\frac{1}{2}[\frac{2E}{3R}] \\ =\frac{E}{3R} \end{gathered}$$
For the branch containing D,E,F,G and H
$$\begin{gathered} R_{eq}=R+\frac{R}{3}+R=\frac{7R}{3} \\ i=\frac{3E}{7R} \end{gathered}$$
Current through D and H$=\frac{3E}{7R}$
Current through E,F,G$=\frac{1}{3}(\frac{3E}{7R})=\frac{E}{7R}$
Hence the bulbs from brightest to dimmest can be arranged as
A > D=H > B=C > E=F=G
Hence, option (a) is correct.