Given,

$\overrightarrow{a}=\hat{i}-\hat{j}+2\hat{k}$ , $\overrightarrow{a}\times \overrightarrow{b}=2\hat{i}-\hat{k}$ and $\overrightarrow{a}·\overrightarrow{b}=3$

Now using the formula ${\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}·\overrightarrow{b}\right|}^2=\left({\left|\overrightarrow{a}\right|}^2·{\left|\overrightarrow{b}\right|}^2{\sin }^2\theta +{\left|\overrightarrow{a}\right|}^2·{\left|\overrightarrow{b}\right|}^2{\cos }^2\theta \right)$

$\Rightarrow {\left|\overrightarrow{a}\times \overrightarrow{b}\right|}^2+{\left|\overrightarrow{a}·\overrightarrow{b}\right|}^2={\left|\overrightarrow{a}\right|}^2·{\left|\overrightarrow{b}\right|}^2\left({\sin }^2\theta +{\cos }^2\theta \right)={\left|\overrightarrow{a}\right|}^2·{\left|\overrightarrow{b}\right|}^2$

We get,

$\Rightarrow {\left|2\hat{i}-\hat{k}\right|}^2+{\left|3\right|}^2={\left|\hat{i}-\hat{j}+2\hat{k}\right|}^2{\left|\overrightarrow{b}\right|}^2$

$\Rightarrow {\left|\overrightarrow{b}\right|}^2=\frac{5+9}{6}=\frac{14}{6}=\frac{7}{3}$

Now finding $\left|\overrightarrow{a}-\overrightarrow{b}\right|=\sqrt{{\left|\overrightarrow{a}\right|}^2+{\left|\overrightarrow{b}\right|}^2-2\left|\overrightarrow{a}\right|\left|\overrightarrow{b}\right|}=\sqrt{6+\frac{7}{3}-6}=\sqrt{\frac{7}{3}}$

Now projection of $\overrightarrow{b}$ on $\overrightarrow{a}-\overrightarrow{b}=\frac{\overrightarrow{b}·\left(\overrightarrow{a}-\overrightarrow{b}\right)}{\left|\overrightarrow{a}-\overrightarrow{b}\right|}$

$=\frac{\overrightarrow{b}·\overrightarrow{a}-{\left|\overrightarrow{b}\right|}^2}{\left|\overrightarrow{a}-\overrightarrow{b}\right|}=\frac{3-\frac{7}{3}}{\sqrt{\frac{7}{3}}}$

$=\frac{2}{\sqrt{21}}$