Given binomial is $\left[x+x^{\log _{10} x}\right]^{5}$
Also given, $T_{3}=10^{6}$
Put $\log _{10} x = y$, the given expression becomes $\left( x + x ^{y}\right)^{5}$.
$T _{3}={ }^{5} C _{2} \cdot x ^{3}\left( x ^{y}\right)^{2}=10 x ^{3+2 y }=10^{6} \ldots .($ given $)$
$\Rightarrow (3+2 y) \log _{10} x=5 \log _{10} 10=5$
$\Rightarrow (3+2 y ) y =5$
$\Rightarrow y =1,-\frac{5}{2}$
$\Rightarrow \log _{10} x =1$ or $\log _{10} x =-\frac{5}{2}$
Therefore, $x =10$ or $x =(10)^{-5 / 2}$