1. For a projectile fired with speed $u$ at angle $\theta$, the horizontal range is

$R=\frac{u^2\sin 2\theta}{g}$

Since the shell reaches the same target at distance $R$ in two possible ways, the two projection angles are complementary:

$\theta_1=\theta,\qquad \theta_2=90^\circ-\theta$

because

$\sin 2\theta=\sin(180^\circ-2\theta).$

2. Time of flight for a projectile is

$T=\frac{2u\sin\theta}{g}$

So the two times are

$t_1=\frac{2u\sin\theta}{g},\qquad t_2=\frac{2u\sin(90^\circ-\theta)}{g}=\frac{2u\cos\theta}{g}.$

3. Their product is

$t_1t_2=\frac{2u\sin\theta}{g}\cdot \frac{2u\cos\theta}{g}$

$t_1t_2=\frac{4u^2\sin\theta\cos\theta}{g^2}$

Using $2\sin\theta\cos\theta=\sin 2\theta$,

$t_1t_2=\frac{2u^2\sin 2\theta}{g^2}.$

But from the range formula,

$R=\frac{u^2\sin 2\theta}{g} \Rightarrow u^2\sin 2\theta=Rg.$

Substitute this into the expression for $t_1t_2$:

$t_1t_2=\frac{2Rg}{g^2}=\frac{2R}{g}.$

4. Therefore, the correct option is

$\boxed{\frac{2R}{g}}$

So, Option A is correct.