How to calculate the gain of the OP-AMP in this configuration
If we make use of the equations for inverting amplifier, then we can see that \$V_{out}=-\frac{R_{f}}{R_{in}}V_{in}\$.

Amplification = \$\frac{V_{out}}{V_{in}}=-\frac{R_{f}}{R_{in}}\$, I will work in absolute, so amplification \$= |-\frac{R_{f}}{R_{in}}|=\frac{R_{f}}{R_{in}}\$

As @Transistor pointed out, OA1 IN- is a virtual earth, this means that \$R_3\$ is in parallel with \$R_2\$. 

If we look back to the expression above and substitute we can understand that 


\$R_{in}\$ for OA1 is \$R_1+R_2//R_3≃504.65\text{ kΩ}\$
\$R_f\$ for OA1 is \$R_4=6.8\text{ kΩ}\$
\$R_{in}\$ for OA2 is \$R_5=4.7\text{ kΩ}\$
\$R_f\$ for OA2 is \$R_6=6.8\text{ kΩ}\$




Amplification for OA1, is \$0.0134≃-37.4\text{ dB}\$, the gain is less than 1.

Amplification for OA2 is \$1.44≃3.2\text{ dB}\$.

Since everything is cascaded, we can just multiply the amplifications, logarithmic addition is the same as linear multiplication. 

Amplification from \$V_{in}\$ to \$V_{out}\$ = \$\frac{V_{out}}{V_{in}}=0.0134×1.44 = -37.4\text{ dB}+3.2\text{ dB}\$



That's \$-34.2\text{ dB}\$, not that good. Or \$0.019296\$. This means that if \$V_{in}=\frac{1}{0.019296}≃52\text{ V}\$, then you will read \$1\text{ V}\$ at the output.