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

The given circuit can be redrawn as

Since, $\dfrac{RAB}{RBC} = \dfrac{RAD}{RDC} = \dfrac{1}{2},$ hence, it is a balanced Wheatstone bridge, therefore current flowing in arm $BD = 0$

$∴ {R}_{eq} = \left(10 + 20\right) ∥ \left(6 + 12\right) = 30 ∥ 18$

$= \dfrac{30 \times 18}{30 + 18} = 11. 25Ω$

$∴ I = \dfrac{V}{Req} = \dfrac{10}{11. 25} = 0. 89A = 0. 9A$