Opamp gain calculation
Well, let's solve and show this mathematically. We are trying to analyze the following circuit (assuming an ideal opamp):







simulate this circuit &ndash; Schematic created using CircuitLab

When we use and apply KCL, we can write the following set of equations:

$$
\begin{cases}
\text{I}_{13}=\text{I}_1+\text{I}_2\\
\\
\text{I}_{13}=\text{I}_3+\text{I}_4\\
\\
\text{I}_5=\text{I}_{13}+\text{I}_{11}\\
\\
\text{I}_5=\text{I}_6\\
\\
\text{I}_7=\text{I}_9\\
\\
\text{I}_{10}=\text{I}_8+\text{I}_9
\end{cases}\tag1
$$

When we use and apply Ohm's law, we can write the following set of equations:

$$
\begin{cases}
\text{I}_2=\frac{\text{V}_\text{c}-\text{V}_1}{\text{R}_2}\\
\\
\text{I}_3=\frac{\text{V}_1-\text{V}_2}{\text{R}_3}\\
\\
\text{I}_4=\frac{\text{V}_1-\text{V}_2}{\text{R}_4}\\
\\
\text{I}_5=\frac{\text{V}_2-\text{V}_3}{\text{R}_5}\\
\\
\text{I}_6=\frac{\text{V}_3}{\text{R}_6}\\
\\
\text{I}_7=\frac{\text{V}_\text{d}-\text{V}_4}{\text{R}_7}\\
\\
\text{I}_8=\frac{\text{V}_5-\text{V}_6}{\text{R}_8}\\
\\
\text{I}_9=\frac{\text{V}_4-\text{V}_6}{\text{R}_9}\\
\\
\text{I}_{10}=\frac{\text{V}_6}{\text{R}_{10}}
\end{cases}\tag2
$$

Now, when we have an ideal opamp we know that \$\text{V}_\text{k}:=\text{V}_{+_1}=\text{V}_{-_1}=\text{V}_1=\text{V}_\text{a}\$ and \$\text{V}_\text{n}:=\text{V}_{+_2}=\text{V}_{-_2}=\text{V}_3=\text{V}_4\$.



Now, applying this to your circuit we need to use (from now on I use the lower case letters for the function in the 'complex' s-domain where I used Laplace transform) the fact that the resistor \$\text{R}_3\$ is replaced by a capacitor, so:

$$\text{R}_3=\frac{1}{\text{sC}}\tag6$$

Solving your question using Mathematica, gives:

In[1]:=Vk = (26/10)/s;
I1 = 1/s;
Vc = 2/s;
Vd = (26/10)/s;
R1 = 47;
R2 = 10^6;
R3 = 1/(s*c);
R4 = 147;
R5 = 200000;
R6 = 10^6;
R7 = 180000;
R8 = 100;
R9 = 10^6;
R10 = 22000;
c = 220*10^(-9);
FullSimplify[
 Solve[{I13 == I1 + I2, I13 == I3 + I4, I5 == I11 + I13, I5 == I6, 
   I7 == I9, I10 == I8 + I9, I1 == (V7 - Vk)/R1, I2 == (Vc - Vk)/R2, 
   I3 == (Vk - V2)/R3, I4 == (Vk - V2)/R4, I5 == (V2 - Vn)/R5, 
   I6 == (Vn)/R6, I7 == (Vd - Vn)/R7, I8 == (V5 - V6)/R8, 
   I9 == (Vn - V6)/R9, I10 == (V6)/R10}, {I2, I3, I4, I5, I6, I7, I8, 
   I9, I10, I11, I13, V2, Vn, V5, V6, V7}]]

Out[1]={{I2 -&gt; -(3/(5000000 s)), 
  I3 -&gt; 8084995149/(5000000 (50000000 + 1617 s)), 
  I4 -&gt; 49999970/(s (50000000 + 1617 s)), 
  I5 -&gt; (-36099977950 + 21021 s)/(6000000 s (50000000 + 1617 s)), 
  I6 -&gt; (-36099977950 + 21021 s)/(6000000 s (50000000 + 1617 s)), 
  I7 -&gt; (184399889750 + 21021 s)/(5400000 s (50000000 + 1617 s)), 
  I8 -&gt; -((110473333739750 + 1282281 s)/(
    59400000 s (50000000 + 1617 s))), 
  I9 -&gt; (184399889750 + 21021 s)/(5400000 s (50000000 + 1617 s)), 
  I10 -&gt; -((2168898699050 + 21021 s)/(1188000 s (50000000 + 1617 s))),
   I11 -&gt; -((1500179599889750 + 48509865789 s)/(
    30000000 s (50000000 + 1617 s))), I13 -&gt; 4999997/(5000000 s), 
  V2 -&gt; -((36099977950 - 21021 s)/(250000000 s + 8085 s^2)), 
  Vn -&gt; -((36099977950 - 21021 s)/(300000000 s + 9702 s^2)), 
  V5 -&gt; -((2663151002587750 + 25834809 s)/(
    66000 s (50000000 + 1617 s))), 
  V6 -&gt; -((2168898699050 + 21021 s)/(2700000000 s + 87318 s^2)), 
  V7 -&gt; 248/(5 s)}}


Which means, for the voltage \$\text{V}_6\$:

V6 -&gt; -((2168898699050 + 21021 s)/(2700000000 s + 87318 s^2))


Which gives: