In the given circuit, the voltmeter reading is $4.5\mathrm{ }\mathrm{V}$ . Assuming that the voltmeter is ideal, current through $12\mathrm{\Omega }$ resistance is -

The voltmeter is ideal so it doesn't draw any current from the circuit.
Potential difference across $9\mathrm{ }\mathrm{\Omega }$ resistance = $4.5\mathrm{ }\mathrm{V}$ (given)

So  the current in the $9 \mathrm{\Omega }$ resistance = $\frac{4.5}{9}=0.5\mathrm{ }\mathrm{A}$

i.e., ${\mathrm{I}}_1^{\mathrm{\prime }}=0.5\mathrm{ }\mathrm{A}$

The same current ${\mathrm{I}}_1^{\mathrm{\prime }}$ passes through $3 \mathrm{\Omega }$. The resistors $9\mathrm{ }\mathrm{\Omega }$ and $3\mathrm{ }\mathrm{\Omega }$ are in series and their equivalent, i.e., $12 \mathrm{\Omega }$ is in parallel with $6\mathrm{ }\mathrm{\Omega }$ between $\mathrm{A}$ and $\mathrm{B}$ . Dividing the current in the inverse ratio of resistances between $\mathrm{A}$ and $\mathrm{B}$ ,

$\frac{{\mathrm{I}}_1^{\mathrm{\prime }}}{{\mathrm{I}}_1^{\mathrm{\prime }\mathrm{\prime }}}=\frac{6}{12}=\frac{1}{2}$

${\mathrm{I}}_1^{\mathrm{\prime }\mathrm{\prime }}=2{\mathrm{I}}_1^{\mathrm{\prime }}=2\mathrm{ }\times \mathrm{ }0.5=1\mathrm{ }\mathrm{A}$

and ${\mathrm{I}}_1={\mathrm{I}}_1^{\mathrm{\prime }}+{\mathrm{I}}_1^{\mathrm{\prime }\mathrm{\prime }}=0.5+1=1.5\mathrm{ }\mathrm{A}$

at junction $\mathrm{C},\mathrm{ }{\mathrm{I}}_1$ divides into three parts. Since the resistances $10\mathrm{\Omega },\mathrm{ }12\mathrm{\Omega },\mathrm{ }15\mathrm{ }\mathrm{\Omega }$ are in parallel between $\mathrm{C}$ and $\mathrm{D}$ , current will distribute in the inverse ratio of resistances.

$\therefore \mathrm{ }{\mathrm{I}}_2^{\mathrm{\prime }}\mathrm{ }:\mathrm{ }{\mathrm{I}}_2^{\mathrm{\prime }\mathrm{\prime }}\mathrm{ }:\mathrm{ }{\mathrm{I}}_2^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=\frac{1}{10}\mathrm{ }:\mathrm{ }\frac{1}{12}\mathrm{ }:\mathrm{ }\frac{1}{15}$

$=6\mathrm{ }:\mathrm{ }5\mathrm{ }:\mathrm{ }4$

${\mathrm{I}}_2^{\mathrm{\prime }}=\frac{5}{15}\times \mathrm{ }1.5=0.5\mathrm{ }\mathrm{A}$

So ${\mathrm{I}}_2^{\mathrm{\prime }}=6\mathrm{k},\mathrm{ }{\mathrm{I}}_2^{\mathrm{\prime }\mathrm{\prime }}=5\mathrm{k},\mathrm{ }{\mathrm{I}}_2^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=4\mathrm{k}$

( $\mathrm{k}$ being a constant of proportionality)

and ${\mathrm{I}}_1={\mathrm{I}}_2^{\mathrm{\prime }}+{\mathrm{I}}_2^{\mathrm{\prime }\mathrm{\prime }}+{\mathrm{I}}_2^{\mathrm{\prime }\mathrm{\prime }\mathrm{\prime }}=15\mathrm{k}$

but ${\mathrm{I}}_1=1.5\mathrm{ }\mathrm{A}$

$\therefore \mathrm{ }15\mathrm{ }\mathrm{k}=1.5$

or $\mathrm{k}=0.1$

so ${\mathrm{I}}_2^{\mathrm{\prime }\mathrm{\prime }}=5\mathrm{k}=0.5\mathrm{ }\mathrm{A}$

Thus current through $12 \mathrm{\Omega }$ resistance is $0.5\mathrm{ }\mathrm{A}$