Let the speed of working of $A$ be $x$ that of $B$ be $y$ and that of $C$ be $z.$
Let the total amount of work be $X$
$\dfrac{0.37X}{x+y+z} = 3\to (1) \qquad\dfrac{0.63X}{x+y}=7 \to (2)$
$y = \frac{5}{4}x (B$ is faster than $A)$
Substituting $y$ in $(2)$ we get $\dfrac{0.63X}{\frac{9}{4}x} = 7 \implies x = 0.04X \implies y = 0.05X $
From $(1)$ we get $\dfrac{0.37X}{0.09X+z} = 3 \implies 0.37X =0.27X+ 3z \implies 3z = 0.1 X \implies z = 0.033 X $
So, $z < x < y$ and $B$ is the fastest worker with working speed of $0.05X$ per day.
So, total number of days required by $B$ = $\dfrac{X}{0.05X} =20.$