If the distance between sun and earth were reduced to half its present value, then the number of days in one year would have been

Correct option is B

​$$\textbf{Concept Used: Kepler’s Third Law of Planetary Motion} \\\text{According to Kepler’s Third Law:} \\T^2 \propto r^3 \\\text{That is, the \textbf{square of the orbital period} of a planet is \textbf{directly proportional} to the \textbf{cube of the radius} (or semi-major axis) of its orbit.}\\[10pt]\textbf{Given:} \\\bullet \text{Initial radius, } r_1, \text{ and time period } T_1 = 365\, \text{days} \\\bullet \text{New radius, } r_2 = \frac{1}{2}r_1\\[10pt]\textbf{Formula Used:} \\\frac{T_2^2}{T_1^2} = \frac{r_2^3}{r_1^3} = \left( \frac{1}{2} \right)^3 = \frac{1}{8}\\[10pt]T_2^2 = \frac{T_1^2}{8} = \frac{365^2}{8} \Rightarrow T_2 = \frac{365}{\sqrt{8}} \approx \frac{365}{2.828} \approx 129.05\, \text{days}$$​​