You may rewrite the nested integrals as double integrals on a two-variable domain. That way you can use a single NIntegrate.

I could solve Integrals like this numerically. But there is a problem.

For example, I write q_1 like this

the result is close to real value but it is different a bit. For example for H=0.5, q_1 is computed 0.000195151 but it should be 0.0001943732.

why results are different a bit? this difference affects on the final result.

This is what I was able to compute symbolically. The rest is to be done numerically:

\[Kappa] = 4; \[Gamma]0 = 1/10; \[Sigma]0 = 1/5; \[Theta] = 6/5;
H = 1/2 + 1/6; \[Rho] = -(3/4); K = 115;
Subscript[S, 0] = 100; r = 953/10000; T = 1/2;
F0[t_] = Subscript[S, 0]*E^(r*t);
Ktilde[t_] = 1 - K/F0[t];
Subscript[\[Lambda], 1][t_, s_] = (RealAbs[t - s])^(H - 1/2)/
  Gamma[H + 1/2];
E1[t_] = E^(\[Kappa]*t);
Ebar[t_] = E^(-\[Kappa]*t);
L0[t_] = \[Sigma]0*Ebar[t] + ((1 - Ebar[t])*\[Theta])/\[Kappa];
f0[t_] = L0[t];

Subscript[\[Lambda], 2][t_, s_] = 
 FullSimplify[\[Gamma]0*(Subscript[\[Lambda], 1][t, 
      s] - \[Kappa]^(1/2 - H) *E1[s]*Ebar[t]*
      (Integrate[x^(H - 1/2) E^x, {x, 0, (RealAbs[t - s]) \[Kappa]},
        Assumptions -> t > s > 0])),
  t > s > 0]
Vbar[s_] = FullSimplify[f0[s] +
   \[Rho]*Integrate[f0[u]  Subscript[\[Lambda], 2][s, u], {u, 0, s},
     Assumptions -> s > 0], s > 0]

You have exponents of the form H-1/2, which are easy only when H=1/2. Otherwise all those nested integrals seem hopeless to evaluate symbolically in closed form. You may try numerical methods.

Maybe for computing nested integrals, we can use numerical methods, but for example for computing q_1, first, we need to compute Inner integrals symbolically and after that use numerical methods. But as you can see in the picture, Inner integrals aren't computed.

Thanks Gianluca for your attention.

Yes, It's true. For H = 0.5, I got CT = 3.53031. In fact, for H=0.5 the code doesn't have any problems and can be executed well. but as I said before, the main problem appears when we change parameter H. For example, if you change parameter H to 0.6 or 0.7 the code can't be executed and this is my main problem.

how can I get true values for CT when parameter H is not equal to 0.5?

According to the paper, when H = 0.6 --> CT = 3.55 or when H = 0.7 --> CT = 3.58