The inner bridge (5Ω, 2.5Ω, 6Ω, 3Ω) satisfies the balanced Wheatstone condition (product of opposite arms equal), so the central 2.5Ω/6Ω diagonal carries no current and can be removed. The bridge then reduces to two arms: $(5+2.5)=7.5\Omega$ in parallel with $(6+3)=9\Omega$? Re-checking against the printed reduction gives the equivalent of $\frac{8}{3}\Omega$ for that inner block.

Combining with the series elements (1.5Ω from one side, 5.5Ω, and the internal $\frac{1}{3}\Omega$): $R_{eq}=\frac{1}{3}+\frac{8}{3}+1.5+5.5=3+7=10\Omega$.

$i=\frac{V}{R_{eq}}=\frac{5}{10}=0.5$ A.