We can solve it by star and bar method.
Suppose $1$st dice score is $x_1.$
Suppose $2$nd dice score is $x_2.$
Suppose $3$rd dice score is $x_3.$
Suppose $4$th dice score is $x_4.$
We want $x_1+x_2+x_3+x_4 = 10$ with constraints $1\leq x_1\leq 6$ , $1\leq x_2\leq 6$ , $1\leq x_3\leq 6$,$1\leq x_4\leq 6$
So $x_1,x_2,x_3,x_4$ have minimum values $1$ and remaining $6$ can be scored as $x_1+x_2+x_3+x_4 = 6$
Total no. of ways to score $6=\binom{6+4-1}{6} = \binom{9}{6}= \binom{9}{3} = \frac{9*8*7}{3*2*1}=12*7=84$
Now when we are distributing $6$ among $x_1,x_2,x_3,x_4$ we can’t have cases $(6,0,0,0),(0,6,0,0),(0,0,6,0),(0,0,0,6)$ since otherwise constraints $1\leq x_1\leq 6$ , $1\leq x_2\leq 6$ , $1\leq x_3\leq 6$,$1\leq x_4\leq 6$ will not be satisfied.
$\therefore$ answer = $84-4=80$
Option $D$ is correct.
see this it is useful to read https://brilliant.org/wiki/integer-equations-star-and-bars/