10. B. Let the four-digit number 
$\overline{a b c d}$ satisfy
$$
a^{3}+b^{3}+c^{3}+d^{3}+1=10 c+d .
$$

Then the number of such four-digit numbers is $\qquad$
10. B. 5 .

From $d^{3} \geqslant d$ we know $c^{3}+1 \leqslant 10 c \Rightarrow 1 \leqslant c \leqslant 3$.
If $c=3$, then $a^{3}+b^{3}+d^{3}=2+d$.
Thus, $d=1$ or 0. Therefore, $a=b=1$, which gives us $1131, 1130$ as solutions.
If $c=2$, then $a^{3}+b^{3}+d^{3}=11+d$. Thus, $d \leqslant 2$.
When $d=2$, $a^{3}+b^{3}=5$, no solution;
When $d=1$ or 0, $a^{3}+b^{3}=11$, no solution.
If $c=1$, then $a^{3}+b^{3}+d^{3}=8+d$. Thus, $d \leqslant 2$.
When $d=2$, $a=b=1$, which gives us $1112$ as a solution;

When $d=1$ or 0, $a=2, b=0$, which gives us $2011, 2010$ as solutions.
In summary, there are 5 four-digit numbers that satisfy the conditions.