1. Given equations

We are given: $\log_{10}(\sin x)+\log_{10}(\cos x)=-1$ for $x\in \left(0,\frac{\pi}{2}\right)$.

Using the log property, $\log_{10}(\sin x\cos x)=-1$ So, $\sin x\cos x=10^{-1}=\frac{1}{10}$

2. Find $\sin x+\cos x$

Use the identity: $(\sin x+\cos x)^2=\sin^2 x+\cos^2 x+2\sin x\cos x$ Since $\sin^2 x+\cos^2 x=1$ and $\sin x\cos x=\frac{1}{10},$ we get $(\sin x+\cos x)^2=1+2\cdot \frac{1}{10}=1+\frac{1}{5}=\frac{6}{5}$

Because $x\in\left(0,\frac{\pi}{2}\right)$, both $\sin x$ and $\cos x$ are positive, hence $\sin x+\cos x=\sqrt{\frac{6}{5}}$

3. Use the second given condition

We are also given: $\log_{10}(\sin x+\cos x)=\frac{1}{2}(\log_{10} n-1)$

Substitute $\sin x+\cos x=\sqrt{\frac{6}{5}}$: $\log_{10}\sqrt{\frac{6}{5}}=\frac{1}{2}(\log_{10} n-1)$

Now, $\log_{10}\sqrt{\frac{6}{5}}=\frac{1}{2}\log_{10}\left(\frac{6}{5}\right)$ So, $\frac{1}{2}\log_{10}\left(\frac{6}{5}\right)=\frac{1}{2}(\log_{10} n-1)$ Multiply both sides by $2$: $\log_{10}\left(\frac{6}{5}\right)=\log_{10} n-1$

Since $1=\log_{10}10$, this becomes $\log_{10}\left(\frac{6}{5}\right)=\log_{10} n-\log_{10}10$ $\log_{10}\left(\frac{6}{5}\right)=\log_{10}\left(\frac{n}{10}\right)$ Therefore, $\frac{n}{10}=\frac{6}{5}$ $n=12$

4. Check options

The correct option is: $\boxed{12}$ which is Option C.