M₁ = initial mass of sun = M
M₂ = final mass of sun = 4 M
T₁ = initial time period
T₂ = final time period
r₁ = initial distance from sun = r
r₂ = final distance from sun = r
Using Kepler's third law
T₁² = 4π²r₁³/(GM₁) eq-1
Using Kepler's third law
T₂² = 4π²r₂³/(GM₂) eq-2
Dividing eq-1 by eq-2
T₁² /T₂² = (4π²r₁³/(GM₁))/(4π²r₂³/(GM₂))
T₁² /T₂² = (M₂/M₁) (r₁³/r₂³)
T₁² /T₂² = (4M/M) (r³/r³)
T₁² /T₂² = 4
T²₁ = 4 T²₂
Taking square root both side
T₁ = 2 T₂
T₂ = T₁ /2
hence the correct choice is
A. 1/2 as long as the present year