I know this problem and the answer the book gives, and unfortunately the book is in error. Let me explain. Adding 0.010 mol of Cr(NO3)3 will also give you 0.010 mol of $\ce{Cr^3+}$, as we can assume the $\ce{Cr^3+}$ will completely dissociate from the $\ce{NO^3-}$. The $\ce{OH^-}$ concentration can be found from the pH, and it is 0.0001 $\frac{moles}{L}$.
The "before reaction starts" gets set up accordingly:
$\ce{Cr^3+ + 4OH- <=> Cr(OH)4-}$
0.010 M .0001 M 0 M
Notice, however, that there is a 4:1 ratio of $\ce{OH^-}$ to $\ce{Cr^3+}$. It takes 4 $\ce{OH^-}$ for every 1 $\ce{Cr^3+}$, and $\ce{OH^-}$ is quite obviously the limiting reactant. So after the reaction, what is the concentration of $\ce{Cr^3+}$? Let's make it 1:1, then.
$\frac{.0001 M}{4}$ = $\ce{2.5 x 10^{-5}}$. And this is essentially the [$\ce{Cr^3+}$] and [$\ce{Cr(OH)4-}$]. However, this is not what the book says. They meant to put the concentration of $\ce{Cr(OH)4-}$ as .010 M, not $\ce{Cr^3+}$. Let's try it that way, and see what we get!
Cr(OH)4- <=> Cr3+ + 4OH-
Here, in the equation, the concentrations are:
$\ce{Cr(OH)4-}$ = .010 M
$\ce{Cr^3+}$ = 0
4$\ce{OH^-}$ = .0001 M
-x +x +4x
.010 - x +x .0001 + 4x
Because .010 - x and .0001 + 4x are so small, let .010 - x = .010 and .0001 + 4x = .0001.
The $\ce{K_f}$ = 8 x $10^{29}$, but that is for the equation when it looks like:
$$\ce{Cr(OH)4- <=> Cr3+ + 4OH-}$$
We switched it around to find out how much $\ce{Cr(OH)4-}$ would dissolve to make $\ce{Cr^3+}$, so we have to use $\frac{1}{Kf}$, which = 1.25 x $10^{-30}$.
$\ce{K_f}$ = $\frac{[{Cr^3+}][{OH-}]^4}{[\ce{Cr(OH)4-}]}$
$\ce{K_f}$ = $\frac{(x)(.0001)^4}{(.010)}$
1.25 x $10^{-30}$ = $\frac{1 x 10^{-16}x}{.010}$
You can take it from there. The answer will be 1.25 x $10^{-16}$ M.