Answer
$F = 280~N$
Work Step by Step
Note that before the window breaks, the system is in equilibrium.
We can find the height where the ladder meets the window:
$h = \sqrt{(5.0~m)^2-(2.5~m)^2} = 4.33~m$
We can choose the base of the ladder as the rotation axis and use the sum of torques to find the force on the ladder from the window:
$\sum \tau_i = 0$
$(1.5~m)(75~kg)(9.8~m/s^2)+(1.25~m)(10~kg)(9.8~m/s^2)-(4.33~m)F = 0$
$(1.5~m)(75~kg)(9.8~m/s^2)+(1.25~m)(10~kg)(9.8~m/s^2)=(4.33~m)F$
$F = \frac{(1.5~m)(75~kg)(9.8~m/s^2)+(1.25~m)(10~kg)(9.8~m/s^2)}{4.33~m}$
$F = 280~N$
