I'm Trying to solve the sde for $X_t^{2}$
I've started with setting $f(x) = X^{2}$. Hence f'(x)=2X and f''(x)=2. $$dY_t= 2X_t(dX_t) + (1/2)2(dX_t)^{2}$$ $$dY_t=2X_t(X_tdB_t)+(X_t)^{2}dt$$ $$dY_t=2X_t^{2}dBt+(X_t)^{2}dt$$ $$Y_t=2\int_0^t(B_s)^{2}dB_s+\int_0^tds$$
Not sure if I've done it right so far and how to go on from here. And if wanted to get the expectation I would typically take the E[itos integral]=0, but im not sure how I could get to that from here.
Your equation seems to be $dX_t=X_t\,dB_t$. It does not have a solution $X_t=B_t$. Your Ito formula up to this point is correct, the conclusion should be $$ dY_t=Y_t\,dt+2Y_t\,dB_t. $$ As $X_t$ is a geometric Brownian motion, its square is one as well, the SDE for $Y_t$, as well as for any other power, thus is expected to have the general form for a geometric BM.