Given :
$w=0.2g,W=20g,\Delta T=0.45$
$\Delta T =\large\frac{1000\times K_f\times w}{m\times W}$
$0.45=\large\frac{1000\times 5.12\times 0.2}{20\times m}$
$\therefore m$(observed)=113.78
Now for $2CH_3COOH \leftrightharpoons (CH_3COOH)_2$
$\qquad\quad 1\qquad\qquad\qquad\qquad 0$
$\qquad\quad 1-\alpha\qquad\qquad\qquad \large\frac{\alpha}{2}$
Where $\alpha$ is the degree of association
$\therefore \large\frac{m_{normal}}{m_{observed}}=$$1-\alpha+\large\frac{\alpha}{2}$
$\large\frac{60}{113.78}$$=1-\alpha+\large\frac{\alpha}{2}$
$\Rightarrow \alpha=0.945$