= Solution
Let $D$ be the separation of the <Earth> and <Moon>, and let $x$ be the spacecraft's distance from the Earth's centre towards the Moon. On $0<x<D$, the two <gravitational forces> oppose each other. Their equality, by <Newton's law of universal gravitation>, is
$$
\frac{GM_E}{x^2}=\frac{GM_M}{(D-x)^2}.
$$
Taking the positive <square root> gives
$$
\boxed{x=\frac{D\sqrt{M_E}}{\sqrt{M_E}+\sqrt{M_M}}
=\frac{D}{1+\sqrt{M_M/M_E}}.}
$$
The <centre of mass> of the two bodies is at $x_G=DM_M/(M_E+M_M)$ from the Earth's centre. Thus the <gravitational field null between two point masses> lies a distance
$$
\boxed{x-x_G=D\left[\frac{1}{1+\sqrt{M_M/M_E}}
-\frac{M_M}{M_E+M_M}\right]}
$$
from their <centre of mass>, on the Moon's side. With $M_E/M_M=81$, $x=9D/10$ and $x_G=D/82$, so
$$
\boxed{x=3.456\times10^8\,\mathrm m,\qquad
x-x_G=\frac{182}{205}D\simeq3.40917\times10^8\,\mathrm m.}
$$
The point is $3.84\times10^7\,\mathrm m$ from the Moon. It is an instantaneous zero of the physical <gravitational field>, not a co-rotating <Lagrange point>: a spacecraft remaining fixed relative to the Earth-Moon line would need a nonzero <centripetal acceleration>. With zero net <force> it cannot follow that circular co-rotation.
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