Answer
$\approx -36~m/s^2$ (other approximations are also possible)
Work Step by Step
Let's find the velocity I reach the ground:
$v^2=v_0^2+2a\Delta x$
where $v_0=0$, $a=g=9.81~m/s^2$, $\Delta x=1.5~m$.
$v^2=0^2+2(9.81~m/s^2)(1.5~m)$
$v=\sqrt {2(9.81~m/s^2)(1.5~m)}=5.42~m/s$
Suppose that it takes $0.15~s$ for me to stop my fall when I reach the ground.
$v=v_0+at$
where $v_0=5.42~m/s$, $v=0$, $t=0.15~s$
$0=5.42~m/s+a(0.10~s)$
$-a(0.10~s)=5.42~m/s$
$a=\frac{5.42~m/s}{-0.15~s}=-36~m/s^2$
