The answer is in part I of Linear Operators by Dunford and Schwartz, Theorem VI.1.4. For any Banach spaces $X$ and $Y$ linear functionals on $\mathcal{B}(X,Y)$ continuous in strong and weak operator topologies are the same. It follows from the proof that they are of the form $F(A)=\sum_{i=1}^n\varphi_i(Ax_i)$ for $x_i\in X$ and $\varphi_i\in Y^*$. In other words, the dual space can be identified with the algebraic tensor product $X\otimes Y^*$, or equivalently the space of finite rank operators $\mathcal{F}(Y,X)$.
For the ultraweak and the ultrastrong topologies on $\mathcal{B}(H)$, where $H$ is a Hilbert space, the answer is in Von Neumann Algebras by Dixmier, Lemma I.3.2. Again, the dual spaces coincide and can be identified with the projective tensor product $H\otimes_\pi H$, or the space of the trace class operators $\mathcal{L}_1(H)$. It also happens to coincide with the norm closure of $\mathcal{F}(H)$ in the Banach dual $\mathcal{B}(H)^*$, and is a Banach predual $\mathcal{B}(H)_*$ of $\mathcal{B}(H)$. Moreover, the ultraweak topology on $\mathcal{B}(H)$ coincides with the weak* topology generated by this predual, they are both $\sigma\big(\mathcal{B}(H),H\otimes_\pi H\big)$.
It doesn't appear that the ultraweak or the ultrastrong topologies were studied much beyond Hilbert spaces, I couldn't even find any established definitions for them. Since the weak operator topology coincides with $\sigma\big(\mathcal{B}(X,Y), X\otimes Y^*\big)$, by analogy to Hilbert spaces a natural candidate for the ultraweak is $\sigma\big(\mathcal{B}(X,Y), X\otimes_\pi Y^*\big)$. However, according to Introduction to Tensor Products of Banach Spaces by Ryan, Section 2.2 a predual to $\mathcal{B}(X,Y)$ is $X\otimes_\pi Y_*$, where $Y_*$ is a predual to $Y$. So the ultraweak so defined does not coincide with the weak* unless $Y$ is reflexive. And even if $Y$ is reflexive it is not clear what a natural candidate for the ultrastrong is or how the duals are related.
