Is it possible to compute $-\log\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right)$ without a calculator? The following question is part of a chemistry problem that came in the Dhaka University admission exam 2013-14.


What is $-\log\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right)$?  
(a) 2.672  
(b) 2.772  
(c) 2.872  
(d) 2.972


My attempt
$$\begin{aligned} -\log_{10}\left({\sqrt{1.8\times 10^{-5}\times 0.1}}\right) &= -\log_{10}\left({\sqrt{1.8\times 10^{-6}}}\right)\\ &\approx -\log_{10}\left({\sqrt{\left(10^{-3}\right)^2}}\right) \\ &=-\log_{10} \left(10^{-3}\right) \\ &= 3 \end{aligned}$$
The options are cruelly close to one another, so the approximation that I have done is of no use to me. How do I find the correct answer by hand quickly (as this is a competitive exam)?

Working with whole numbers
$$1.8\times 10^{-5}\times 0.1= \frac 9 5 \times 10^{-6}$$
$$\sqrt{1.8\times 10^{-5}\times 0.1}=\frac{3}{ \sqrt{5}}\times 10^{-3}$$
$$\frac{3}{ \sqrt{5}}=\frac{3}{ \sqrt{4+1}}=\frac 3 2 \times \frac 1{\sqrt{1+\frac 14}}\sim\frac 3 2 \times \left(1-\frac 18\right)=\frac{21}{16}=1+\frac 5 {16}$$
$$\log_{e}\left(1+\frac 5 {16}\right)\sim\frac 5 {16}\implies \log_{10}\left(1+\frac 5 {16}\right)\sim  \frac {5}{16 \times 2.30}  \sim 0.135$$
Then $-2.865$ by hand
Edit
A good rational approximation of $\log_e(10)$ is  $\frac{76}{33}$
$$\log_{10}\left(1+\frac 5 {16}\right)\sim  \frac {5}{16} \times  \frac{33}{76}=\frac{165}{1216}\sim\frac{165}{1215}=\frac{11}{81}\times  \frac{125}{125}=\frac{1375}{10125} \sim\frac{1375}{10000}=0.1375$$
