If the fourth term in the binomial expansion of

$\therefore$ fourth term is equal to 200 .

$T_{4}={ }^{6} C_{3}\left(\sqrt{x\left(\frac{1}{1+\log _{10} x}\right)}\right)^{3}\left(x^{\frac{1}{12}}\right)^{3}=200$

$\Rightarrow 20 x^{\frac{3}{2\left(1+\log _{10} x\right)}} \cdot x^{\frac{1}{4}}=200$

$x^{\frac{1}{4}+\frac{3}{2\left(1+\log _{10} x\right)}}=10$

Taking $\log _{10}$ on both sides and putting $\log _{10} x=t$

$\left(\frac{1}{4}+\frac{3}{2(1+t)}\right) t=1 \Rightarrow t^{2}+3 t-4=0$

$\Rightarrow t^{2}+4 t-t-4=0 \Rightarrow t(t+4)-1(t+4)=0$

$\Rightarrow t=1$ or $t=-4$

$\log _{10} x=1 \Rightarrow x=10$

or $\log _{10} x=-4 \Rightarrow x=10^{-4}$

According to the question $x>1, \therefore x=10$.