It is possible to project a particle with a given speed in two possible ways so as to have the same range, R. The product of the times taken to reach this point in the two possible ways is proportional to

The correct option is B

R

The range of a projectile is given by $R=\frac{2v_0^{2}cos\theta sin\theta }{g}$.
Now, the range at complementary angles $\theta$ and ${90}^{\circ }-\theta$ is equal. The times of flight coresponding to these two angles are
$t_1=\frac{2v_{0}sin\theta }{g}........\left(1\right)$
and $t_2=\frac{2v_{0}sin({90}^{\circ }-\theta )}{g}=\frac{2v_{0}cos\theta }{g}........\left(2\right)$
From equations (1) and (2), $t_1{t}_2=\frac{4v_0^{2}sin\theta cos\theta }{g^2}=\frac{2R}{g}$
Thus, $t_1{t}_2\propto R$ and the correct choice is (b).