Suppose earth's orbital motion around the sun supposed to be in circular is suddenly stopped, what time will the earth take to fall into the sun?

Kepler's law applies to planetary orbits, whether they be of circular, or elliptical shape. It says that $\frac{T_2^2}{T_2^1}=\frac{R_3^2}{R_3^1}$ where T is the period of an orbit and R is its semi-major axis. The semi-major axis is the average of the planet's maximum and minimum distances from the sun.

Let the earth's mean radius be $R_1$..This straight fall can be considered 1/2 of a degenerate elliptical orbit with major axis equal to  $R_1$. Its semi-major axis is $\frac{R_1}{2}$ (the average of R1 and zero). Its period will be designated $T_2$.

So: $\frac{T_2^2}{T_1^2}=\frac{{\left(\frac{R_1}{2}\right)}^3}{R_1^3}={\left(\frac{1}{2}\right)}^3$

And therefore, $T_2=0.353$ year, and the time to fall into the sun is the time to fall into the sun is ,64.52 days