= Solution
Use four-variable <Karnaugh maps> with rows $AB=00,01,11,10$ and columns $Cx=00,01,11,10$, both in <Gray code> order. A dash denotes an unused present state. The maps for the three next-state bits are
$$
\begin{array}{c|cccc}
A^+&00&01&11&10\\\hline
00&0&0&0&0\\01&0&0&1&0\\11&-&-&-&-\\10&0&0&-&-
\end{array}\qquad
\begin{array}{c|cccc}
B^+&00&01&11&10\\\hline
00&0&0&1&0\\01&1&1&0&0\\11&-&-&-&-\\10&0&1&-&-
\end{array}\qquad
\begin{array}{c|cccc}
C^+&00&01&11&10\\\hline
00&0&1&0&0\\01&1&0&0&0\\11&-&-&-&-\\10&0&0&-&-
\end{array}.
$$
For $A^+$, join its sole one with the dash directly below it. For $B^+$, use the four-cell groups $Ax$ and $B\bar C$, and the two-cell group $\bar BCx$; the last group wraps between the top and bottom rows. For $C^+$, join the one at row $01$, column $00$ to the dash below it; the one at row $00$, column $01$ is isolated. These <prime implicants> give
$$
\boxed{D_A=BCx,\qquad D_B=Ax+B\bar C+\bar BCx,\qquad D_C=B\bar C\bar x+\bar A\bar B\bar Cx.}
$$
Here products mean <Boolean conjunction> and sums mean <Boolean disjunction>. The $D$ inputs are the desired next-state bits of the three <D flip-flops>. The equations are independent of the unused values except for the explicitly chosen <Karnaugh map> groups.
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