Answer
$h = 33~m$
Work Step by Step
Let $h$ be the height of the cliff. Let $T$ be the total time it takes until we hear the sound.
Let $t_1$ be the time that the stone drops.
$h = \frac{1}{2}gt_1^2$
Let $t_2$ be the time that the sound of the splash travels up the cliff. Note that $t_1+t_2 = T$
$h = v~t_2$
$h = v~(T-t_1)$
$h = v~T-v~t_1$
We can equate the two expression for $h$ and solve for $t_1$.
$\frac{1}{2}gt_1^2 = v~T-v~t_1$
$\frac{1}{2}gt_1^2 +v~t_1-v~T = 0$
$\frac{1}{2}(9.80~m/s^2)t_1^2 +(343~m/s)~t_1-(343~m/s)(2.7~s) = 0$
$(4.90~m/s^2)t_1^2 +(343~m/s)~t_1-(926.1~m) = 0$
We can use the quadratic formula to find $t_1$.
$t_1 = \frac{-b\pm \sqrt{b^2-4ac}}{2a}$
$t_1 = \frac{-(343)\pm \sqrt{(343)^2-(4)(4.9)(-926.1)}}{(2)(4.9)}$
$t_1 = 2.6~s, -73~s$
Since the negative value is unphysical, the solution is $t_1 = 2.6~s$.
We can find the height of the cliff.
$h = \frac{1}{2}gt_1^2$
$h = \frac{1}{2}(9.8~m/s^2)(2.6~s)^2$
$h = 33~m$