Let $\overrightarrow b=x\hat i+y\hat j+z\hat k$,
Given: $\overrightarrow a.\overrightarrow b=3$
$\Rightarrow\: x+y+z=3$..........(i)
Also given that $\overrightarrow a\times\overrightarrow b=\overrightarrow c$
$\Rightarrow\: (z-y)\hat i+(x-z)\hat j+(y-x)\hat k=\hat j-\hat k$
$\Rightarrow\:z=y,x-z=1,\:and\:y-x=-1$....(ii)
Solving (i) and (ii) we get $x=\large\frac{5}{3}$, $y=\large\frac{2}{3}$ and $z=\large\frac{2}{3}$
$\therefore \overrightarrow b=\large\frac{1}{3}$$(5\hat i+2\hat j+2\hat k)$