Answer
$0.99999986\;\rm m$
Work Step by Step
We know that the two spheres will attract each other, so the distance between them will be less than 1 m.
But the difference will be negligible since the gravitational mass force is weak.
We can see, in the figure below, that the two spheres are in equilibrium after attaching them for a while.
The distance between the center of the sphere is given by
$$x=1-2d$$
where $d$ is given by $\sin\theta=d/L$, so
$$x=1-2L\sin\theta\tag 1$$
Now we need to find the net force exerted on the left sphere, so we can find the distance between them.
$$\sum F_{x,\rm on \;2}=F_{\rm 1\;on\; 2}-T\sin\theta=ma_x=m(0)=0$$
Hence,
$$F_{\rm 1\;on\; 2}=T\sin\theta$$
where $F_{\rm 1\; on\; 2}$ is a gravitational force exerted by sphere 1 on sphere 2.
$$\dfrac{Gm_1m_2}{x^2} =T\sin\theta$$
Noting that $m_1=m_2=m$
$$\dfrac{Gm^2}{x^2} =T\sin\theta\tag 2$$
$$\sum F_{y,\rm on \;2}=T\cos\theta-mg=ma_y=m(0)=0$$
Thus,
$$T\cos\theta=mg$$
Hence,
$$T=\dfrac{mg}{\cos\theta}$$
Plugging into (2);
$$\dfrac{Gm^{ \color{red}{\bf\not}2}}{x^2} =\dfrac{ \color{red}{\bf\not}mg}{\cos\theta}\sin\theta $$
So that
$$\tan\theta=\dfrac{Gm}{x^2g}$$
We know that $x\approx 1.0\;\rm m$ since the gravitational force is too weak as we mentioned above.
Thus,
$$ \theta=\tan^{-1}\left(\dfrac{(6.67\times10^{-11})(100 )}{(1)^2(9.8)}\right)=\bf (3.899\times10^{-8})^\circ $$
Plugging the known into (1);
$$x=1-2(100)\sin(3.899\times10^{-8})^\circ$$
$$x=\color{red}{\bf 0.99999986}\;\rm m\approx 1\;m $$
