Answer
$0.14~kg$
Work Step by Step
$1~mi=1609~m$, $1~h=3600~s$
$\Delta x=0.15~m$, $v=0$, $m=0.15~kg$
$v_0=92~mi/h=(92~mi/h)(\frac{1609~m}{1~mi})(\frac{1h}{3600~s})=41.1~m/s$
$v^2=v_0^2+2a\Delta x$
$0^2=(41.1~m/s)^2+2a(0.15~m)$
$(-0.3~m)a=(41.1~m/s)^2$
$a=\frac{(41.1~m/s)^2}{-0.3~m}=-5630~m/s^2$. It means that acceleration and displacement are in opposite directions.
$F=m|a|$ -> $m=\frac{F}{|a|}=\frac{803~N}{5630~m/s^2}=0.14~kg$
