1. Set up the area integral

The region is bounded by $\frac{a}{x^2} \le y \le \frac{1}{x}, \qquad 1 \le x \le 2, \qquad 0<a<1.$

So the area is $A=\int_1^2 \left(\frac{1}{x}-\frac{a}{x^2}\right)dx.$

Given that $A=\ln 2-\frac{1}{7}.$

2. Evaluate the integral

We compute: $\int_1^2 \frac{1}{x}\,dx=\ln 2$ and $\int_1^2 \frac{1}{x^2}\,dx=\int_1^2 x^{-2}\,dx=\left[-x^{-1}\right]_1^2= -\frac12-(-1)=\frac12.$

Therefore, $A=\ln 2-a\left(\frac12\right)=\ln 2-\frac{a}{2}.$

3. Use the given area

Since $\ln 2-\frac{a}{2}=\ln 2-\frac{1}{7},$ we get $\frac{a}{2}=\frac{1}{7}$ so $a=\frac{2}{7}.$

4. Find $7a-3$

$7a-3=7\cdot \frac{2}{7}-3=2-3=-1.$

5. Match with the options

The value is $-1,$ which corresponds to Option D.