The magnitude of current in the middle branch in ampere unit is

$\begin{array}{l}Voltageis\text{\hspace{0.17em}same in all three branches}\\ \text{60i=20i'=10}\left(1-i-i\text{'}\right)\\ \frac{i}{i\text{'}}=\frac{20}{60}=\frac{1}{3}\\ i\text{'}=3i\\ 6i=1-i-3i\\ i=\frac{1}{10}A\\ i\text{'}=3i=0.3A\end{array}$