By KVL
$$
\begin{aligned}
& 4 \mathrm{i}+8 \mathrm{i}_1-5=0 [In smaller loop]\\
& \Rightarrow 4 \mathrm{i}_1+4 \mathrm{i}_2+8 \mathrm{i}_1-5=0 \\
& \Rightarrow 12 \mathrm{i}_1+4 \mathrm{i}_2=5 ...(i)\\
& \text { and, } 5 \mathrm{i}_2+3-8 \mathrm{i}_1=0 [In bigger loop]\\
& 5 \mathrm{i}_2-8 \mathrm{i}_1=-3 \\
& 5 \mathrm{i}_2-8\left(\frac{5-4 \mathrm{i}_2}{12}\right)=-3 \quad[\text { from }(\mathrm{i})] \\
&
\end{aligned}
$$
$\begin{aligned} & 5 \mathrm{i}_2-\frac{2}{3}\left(5-4 \mathrm{i}_2\right)=-3 \\ & 5 \mathrm{i}_2-\frac{10}{3}+\frac{8}{3} \mathrm{i}_2=-3 \\ & \frac{23}{3} \mathrm{i}_2=\frac{1}{3} \\ & \mathrm{i}_2=\frac{1}{23} \Rightarrow \text { Current through } 5 \Omega=\frac{1}{23} \mathrm{~A}\end{aligned}$