#### Answer

$m = 89.3~kg$

#### Work Step by Step

We can find the acceleration of the system.
$y = \frac{1}{2}at^2$
$a = \frac{2y}{t^2}$
$a = \frac{(2)(1.0~m)}{(6.0~s)^2}$
$a = 0.0556~m/s^2$
Let's consider the system of both blocks. The weight of the block of mass $m$ pulls down on the left while the weight of the 100-kg block pulls down on the right. We can use a force equation to find $m$.
$\sum F = (m+100~kg)~a$
$(100~kg)~g-mg = (m+100~kg)~a$
$m = \frac{(100~kg)(g-a)}{g+a}$
$m = \frac{(100~kg)(9.80~m/s^2-0.0556~m/s^2)}{9.80~m/s^2+0.556~m/s^2}$
$m = 89.3~kg$