Answer
$K = 109~J$
Work Step by Step
We can find the speed at the end of the fall:
$v_f^2 = v_0^2+2ay$
$v_f = \sqrt{v_0^2+2ay}$
$v_f = \sqrt{(3.00~m/s)^2+(2)(9.8~m/s^2)(4.00~m)}$
$v_f = 9.35~m/s$
We can find the kinetic energy:
$K = \frac{1}{2}mv^2$
$K = \frac{1}{2}(2.50~kg)(9.35~m/s)^2$
$K = 109~J$
