The solution of the Black-scholes equation is the price of a European call. And the option price assumes the underlying stock is a geometric Brownian motion with volatility $\sigma_{1}>0$.
Suppose, however, the underlying asset is really a geometric Brownian motion with volatility $\sigma_{2} > \sigma_{1}$, i.e. \begin{equation} dS(t) = \alpha S(t)dt + \sigma_{2}S(t)dW(t). \end{equation}
Consequently, the market price of the call is incorrect.
Can we set up a portfolio which has an arbitrage opportunity in the market? Furthermore, if there any methods to generate a portfolio arbitrage opportunity (how to consider this problem)?
This is an excellent question! You can! This is basically the result called "The fundamental theorem of derivatives trading" see link. We assume $$ dS(t) =S(t) (\alpha dt + \sigma_2 dW(t) ) $$ but some derivative is priced at implied volatility $\sigma_1<\sigma_2$, assume it has payoff function $h(S_T)$ and has value $C_h(S_t,r,\sigma_2,t)=C(t,S_t)$ consider a last $\hat{\sigma}^2$, this is the volatility we will use for hedging and can be either or something else. Assume we have constant interest rate $r$. The derivative price function satisfies the Black-Scholes PDE $$ \frac{\partial C}{\partial t} + rs \frac{\partial C}{\partial s}+ 1/2 \sigma_2^2 s^2 \frac{\partial^2 C}{\partial s^2}- rC = 0 $$ and C(T,s) = h(s). To make the argument similar to usual hedging strategies we will assume we sell the underlying and hedge it - in fact to enjoy the arbitrage we need to exactly reverse the strategy described. Say we sell the derivative and delta hedge it using $\hat{\sigma}$ we will continuously hold $\phi_t = \frac{\partial C}{\partial s}$ (determined by the BS equation with $\hat{\sigma}^2$) of the risky asset and make the hedging portfolio self-financing in the risk free we get $$ dX_t = \frac{\partial C}{\partial s}(S_t) dS_t + (X_t -\frac{\partial C}{\partial s}(S_t) \cdot S_t ) r dt $$ while the actual price process $Y$ (Ito on C from BS satisfies) $$ dY_t = \frac{\partial C}{\partial s} dS_t + (\frac{\partial C}{\partial t}(S_t) +1/2 \sigma_2^2 S_t^2 \frac{\partial^2 C}{\partial s^2}(S_t) )dt $$ so the hedge error $Z = X - Y$ is (use BS equation) $$ dZ_t = r X_t + 1/2 S_t^2 \frac{\partial^2 C}{\partial s^2}(S_t) (\hat{\sigma}^2-\sigma^2_2 ) dt $$ this yields that (solve it as a differential equation) $$ Z_T = X_T-h(S_T) = \int_0^T e^{r(T-s)} 1/2 S_t^2 \Gamma_t ^2 (\hat{\sigma}^2 -\sigma_2 ^2) dt $$ note all terms in the integral are positive so if we hedge using $\hat{\sigma}=\sigma_1 <\sigma_2$ the (combined) position is free but will surely loose money, while in the case $\hat{\sigma}=\sigma_1$ the integral cancels, but the position will initially cost us money (we initiate a replication of a more expensive derivative). Ergo reversing the strategy yields an arbitrage :)