Answer
$\theta=cos^{-1}(\frac{\sqrt {Rg}}{V})$
Work Step by Step
Please see the attached image first.
Let's apply equation $S=ut$ in the horizontal direction of AB motion.
$\rightarrow S=ut+\frac{1}{2}at^{2}$
Let's plug known values into this equation.
$2R=Vcos\theta\space t$
$\frac{2R}{Vcos\theta}= t-(1)$
Let's apply equation $S=ut+\frac{1}{2}at^{2}$ in vertical direction of AB motion.
$\uparrow S=ut+\frac{1}{2}at^{2}$
Let's plug known values into this equation.
$R=Vsin\theta\times t-\frac{1}{2}(-g)t^{2}-(2)$
$(1)=\gt(2),$
$-R=Vsin\theta\times \frac{2R}{Vcos\theta}-\frac{1}{2}(-g)(\frac{2R}{Vcos\theta})^{2} $
$-R= \frac{2Rsin\theta}{cos\theta}-\frac{2gR^{2}}{V^{2}cos^{2}\theta}-(3)$
The problem said that R is the horizontal range the projectile would have had if launched over level ground at the same launch angle. So we previously learned,
$R=\frac{2V^{2}}{g}sin\theta cos\theta=\gt sin\theta=\frac{Rg}{V^{2}cos^{2}\theta}-(4)$
$(4)=\gt(3)$
$-R=(\frac{2R}{cos\theta})(\frac{Rg}{2V^{2}cos\theta})-(\frac{2R^{2}g}{V^{2}cos^{2}\theta})$
$-1=\frac{-Rg}{V^{2}cos^{2}\theta}$
$cos\theta = \sqrt {\frac{Rg}{V^{2}}}=\gt\theta=cos^{-1}(\frac{\sqrt {Rg}}{V})$
