For a balanced Wheatstone's bridge

$\frac{P}{Q}=\frac{R}{S}$

$\therefore \quad \frac{10 \,\Omega}{30\, \Omega}=\frac{30\, \Omega}{90\, \Omega}$ or $\frac{1}{3}=\frac{1}{3}$

It is a balanced Wheatstone's bridge. Hence no current flows in the galvanometer arm. Hence, resistance $50\, \Omega$ becomes ineffective.

$\therefore$ The equivalent resistance of the circuit is

$R_{\mathrm{eq}}=5\, \Omega+\frac{(10 \,\Omega+30\, \Omega)(30\, \Omega+90\, \Omega)}{(10\, \Omega+30\, \Omega)+(30 \,\Omega+90\, \Omega)} $

$=5\, \Omega+\frac{(40\, \Omega)(120\, \Omega)}{40 \,\Omega+120\, \Omega}=5 \,\Omega+30\, \Omega=35\, \Omega$

Current drawn from the cell is

$I=\frac{7 \mathrm{V}}{35\, \Omega}=\frac{1}{5} \mathrm{A}=0.2 \,\mathrm{A}$