An electricity and magnetism problem by Sriram Radhakrishnan

$\large \displaystyle \text{Equivalent resistance, } \frac{1}{R_P} = \frac{1}{3} + \frac{1}{1+2}\\ \large \displaystyle \implies R_P = \frac{3}{2} = 1.5\\ \large \displaystyle V = IR\\ \large \displaystyle \implies I = \frac{V}{R}\\ \large \displaystyle \implies I = \frac{0.5}{1.5} \approx \color{#3D99F6}{\boxed{0.33}}.$

$\begin{aligned} I & = & \frac { V }{ R } \\ \quad & = & \frac { 0.5 }{ \frac { 1 }{ \frac { 1 }{ 3 } +\frac { 1 }{ 1+2 } } } \\ \quad & = & \frac { 1 }{ 3 } \\ \quad & \approx & \boxed { 0.333\space A } \end{aligned}$