The value of current $I$ as shown in the given circuit diagram is

The given circuit can be redrawn as

Since, $\frac{R_{AB}}{R_{BC}}=\frac{R_{AD}}{R_{DC}}=\frac{1}{2},$ hence, it is a balanced Wheatstone bridge, therefore current flowing in arm $BD=0$

$\therefore R_{\mathrm{eq}}=\left(10+20\right)\parallel \left(6+12\right)=30\parallel 18$

$=\frac{30\times 18}{30+18}=11.25 \mathrm{\Omega }$

$\therefore I=\frac{V}{R_{\mathrm{eq}}}=\frac{10}{11.25}=0.89 A=0.9 A$