1. Let $a=\sqrt{x^{\left(\frac{1}{1+\log_{10}x}\right)}}=x^{\frac{1}{2(1+\log_{10}x)}}$ and $b=x^{1/12}.$

Then the given expression is $(a+b)^6.$

2. The 4th term in the expansion of $(a+b)^6$ is $T_4=\binom{6}{3}a^{6-3}b^3=\binom{6}{3}a^3b^3.$ Since $\binom{6}{3}=20$, $T_4=20a^3b^3.$

3. Compute $a^3$ and $b^3$: $a^3=\left(x^{\frac{1}{2(1+\log_{10}x)}}\right)^3=x^{\frac{3}{2(1+\log_{10}x)}},$ $b^3=\left(x^{1/12}\right)^3=x^{1/4}.$ Hence, $T_4=20x^{\frac{3}{2(1+\log_{10}x)}+\frac14}.$

Given $T_4=200$, $20x^{\frac{3}{2(1+\log_{10}x)}+\frac14}=200$ which gives $x^{\frac{3}{2(1+\log_{10}x)}+\frac14}=10.$

4. Put $y=\log_{10}x.$ Then $x=10^y$ with $y>0$ since $x>1$.

Now, $x^{\frac{3}{2(1+y)}+\frac14}=10$ becomes $(10^y)^{\frac{3}{2(1+y)}+\frac14}=10.$ So, $10^{y\left(\frac{3}{2(1+y)}+\frac14\right)}=10.$ Therefore, $y\left(\frac{3}{2(1+y)}+\frac14\right)=1.$

5. Solve this equation: $\frac{3y}{2(1+y)}+\frac{y}{4}=1.$ Multiply by $4(1+y)$: $6y+y(1+y)=4(1+y).$ $6y+y+y^2=4+4y$ $y^2+3y-4=0.$ Factorizing, $(y-1)(y+4)=0.$ So, $y=1 \quad \text{or} \quad y=-4.$ Since $y>0$, we take $y=1.$

6. Thus $\log_{10}x=1 \implies x=10.$

7. Check with options: this is Option C.