Ans is option (C)
For the value inside the square root to be real, it should be greater than or equal to zero.
$\therefore$ log$_{e}\frac{4x-x^{2}}{3}\geqslant0$
$\Rightarrow$ $\frac{4x-x^{2}}{3}\geqslant1$
$\Rightarrow$ $\frac{4x-x^{2}}{3}-1\geqslant0$
$\Rightarrow$ $\frac{4x-x^{2}-3}{3}\geqslant0$
Multiplying by 3 on both sides inequality, we get $4x-x^{2}-3\geqslant0$
Multiplying by -1 on both sides inequality, we get $x^{2}-4x+3\leqslant0$
$\Rightarrow$ $(x-1)(x-3)\leqslant0$ $\Rightarrow$ $x\in[1,3]$