#### Answer

$d = 1.4~m$

#### Work Step by Step

Let the left end of the 1.0-kg object be the position x = 0.
Let $M_1$ be the 1.0-kg object. We can find $r_1$, the distance from the center of mass to the pivot.
$r_1 = d - 1.0 ~m$
Let $M_2$ be the 4.0-kg object. We can find $r_2$, the distance from the center of mass to the pivot.
$r_2 = 1.5~m - d$
Since the two objects are in equilibrium, the magnitude of their torques about the pivot are equal. We can find $d$.
$\tau_1 = \tau_2$
$r_1~ F_1 = r_2~ F_2$
$(d-1.0~m)~ M_1~g = (1.5~m-d)~ M_2~g$
$(d)(M_1+M_2) = (1.5~m)~M_2+(1.0~m)~M_1$
$d = \frac{(1.5~m)~M_2+(1.0~m)~M_1}{M_1+M_2}$
$d = \frac{(1.5~m)(4.0~kg)+(1.0~m)(1.0~kg)}{1.0~kg+4.0~kg}$
$d = 1.4~m$