Answer
At a distance of $\frac{d}{3}$ from the star of mass $M_1$, which is a distance of $\frac{2d}{3}$ from the star of mass $4.0~M_1$, the gravitational field due to the two stars is zero.
Work Step by Step
Let's draw a straight line connecting the centers of the two stars. Let $x$ be the distance to a point on this line from the star with mass $M_1$. Then the distance to the other star is $d-x$.
If the magnitude of the gravitational field from each star is equal at a point along this line, then the gravitational field due to the two stars will be zero, since the gravitational field around each star points in opposite directions for points along this line.
We can equate the magnitudes of the gravitational fields and solve for $x$:
$\frac{G~M_1}{x^2} = \frac{G~(4.0~M_1)}{(d-x)^2}$
$4.0~x^2 = d^2-2dx+x^2$
$3.0~x^2+2dx-d^2 = 0$
$(3.0~x-d)(x+d) = 0$
$x = \frac{d}{3}, -d$
Note that $x=-d$ is inadmissible, because at this point, the gravitational field around each star points in the same direction.
At a distance of $\frac{d}{3}$ from the star of mass $M_1$, which is a distance of $\frac{2d}{3}$ from the star of mass $4.0~M_1$, the gravitational field due to the two stars is zero.
