I = \(\frac{E}{R + r}\)

For parallel of 5Ω and 10Ω
\(\frac{1}{R}\) = \(\frac{1}{10}\) + \(\frac{1}{5}\) = \(\frac{1+2}{10}\) = \(\frac{3}{10}\)
∴R = \(\frac{10}{3}\)
Total R = 3\(\frac{1}{3}\) + 3 x 1 = 7\(\frac{1}{3}\)
∴I = 6\(\div\) 7\(\frac{1}{3}\) = \(\frac{6}{1}\) x \(\frac{3}{22}\) = \(\frac{18}{22}\) = \(\frac{9}{11}\)
∴I = \(\frac{9}{11}\)