A wedge Y with mass of 10 kg and all frictionless surfaces and the...

Horizontal acceleration of the wedge (assuming the block and wedge move together horizontally):
$A = \dfrac{f}{M + m} = \dfrac{24}{10 + 2} = 2\text{ m/s}^2$

Working in the non-inertial frame of the wedge, the block experiences a pseudo force $mA$ directed opposite to the wedge's acceleration. The components along the incline are:
Component of gravity down the incline: $g\sin 37^\circ$
Component of pseudo force up the incline: $A\cos 37^\circ$

Net relative acceleration of the block down the incline:
$a_{\text{rel}} = g\sin 37^\circ - A\cos 37^\circ$

Using $\sin 37^\circ = 0.6$ and $\cos 37^\circ = 0.8$:
$a_{\text{rel}} = 10(0.6) - 2(0.8) = 6 - 1.6 = 4.4\text{ m/s}^2$

Time taken to slide a distance of $8.8\text{ m}$ along the incline, starting from rest:
$s = \dfrac{1}{2} a_{\text{rel}}\, t^2$

$8.8 = \dfrac{1}{2}(4.4)\,t^2$

$8.8 = 2.2\,t^2$

$t^2 = 4 \Rightarrow t = 2\text{ s}$

Hence, the required time is $2\text{ s}$.