Consider the trajectory to be an Ellipse,
Semi-major axis, a=ae+am2=1.89×1011m
Radius of the earth = 1.5×1011m
Now,
We consider a circle whose area is equal to the path covered by ellipse,
∴πR2=πab2
∴R=√ab2
=√1.89×1011×1.5×10112
=1.19×1011m
Applying kepler's law,
∴T2∝r3
∴(T365)2=(Rae)3
∴T2=(1.191.5)3×3.652
∴T≃260days.