Answer
$a \gt b = c$
Work Step by Step
Let $V$ be the potential of battery and $R$ be the resistance of bulb.
Potential across $a$ is $V$ therefore power
$ P_a = \frac{V^2}{R}$
Since $b$ and $c$ are in series therefore potential across each is $\frac{V}{2}$
$P_b = \frac{V^2}{4R}$ and
$P_c =\frac{V^2}{4R}$
As intensity is proportional to power therefore order of brightness is
$a \gt b = c$
