My reasoning for this question is:
As we have a $DFA$ having $3$ states with exactly two as the final states.
so to select $2$ final states from $3$ states we can select it in $^{3}\textrm{C}_{2}$ ways so now we have $3$ $DFA’s$
The alphabet given in the question contains $3$ symbols $({0,1,2})$
$(each$ $symbol$ $from$ $each$ $state$ $has$ $3$ $choices$ $either$ $it$ $can$ $loop$ $over$ $the$ $same$ $state$, $or$ $go$ $to$ $any$ $one$ $state$ $from$ $the$ $remaining$ $two$)
And , as it is a $DFA$ every state should have transitions for all the symbols and hence for $3$ states each having $3$ symbols . (For forming a valid String we are only taking transition on one symbol per state).
thus the choices we have for all the $3$ states and $3$ symbols will be $3^9$
and initially we had $^{3}\textrm{C}_{2}$ for selecting the final states thus total $DFA’s$ possible
$3^9*3$
$3^{10}$
