Answer: B

Small Hint:

Use radii $\frac{3}{2},$ $1,$ and $\frac{1}{2},$ and let a new gauge have radius $r$

Big Hint:

If its center is $(u,v)$, subtract its three tangency-distance equations

Solution:

Place the hole’s center at the origin. The $2$-inch and $1$-inch plug centers are $(0,-\frac{1}{2})$ and $(0,1).$ Let a new gauge have radius $r$ and center $(u,v).$ Tangency to the two plugs gives $$\begin{gathered} u^2+(v+\frac{1}{2})^2=(1+r)^2,\\ u^2+(v-1)^2=(\frac{1}{2}+r)^2, \end{gathered}$$ while internal tangency to the hole gives $u^2+v^2=(\frac{3}{2}-r)^2.$ Subtracting pairs of equations yields $v=\frac12+\frac r3$ and $v=\frac32-2r.$ Thus $r=\frac{3}{7},$ so $d=2r=\frac{6}{7}\approx0.86.$

Therefore, the correct answer is B.