Let the potential at $O$ is $V_{0}$.
Application of Kirchhoff's first law at junction $O$ gives
$\frac{8-V_{0}}{2} =\frac{V_{0}-4}{4}+\frac{V_{0}-2}{2}$
$=\frac{V_{0}-4+2 V_{0}-4}{4} $
$\frac{4\left(8-V_{0}\right)}{2} =V_{0}-4+2 V_{0}-4$
$16-2 V_{0} =3 V_{0}-8 $
$16+8 =5 V_{0} $
$V_{0} =\frac{24}{5}=4.8\, V$
