Finding Cohomology of a Fiber Bundle from Serre Spectral Sequence

Question

I have a fiber bundle $F\rightarrow E\rightarrow \mathbb{RP}^3$ and I know the cohomology groups of $F$ and $\mathbb{RP}^3,$ $H^0(F,\mathbb Z)=H^3(F,\mathbb Z)=\mathbb Z$, $H^2(F,\mathbb Z)=(\mathbb Z/2\mathbb Z)^2$ and other cohomology groups of $F$ are zero. Additionally, $H^0(\mathbb{RP}^3,\mathbb Z)=H^3(\mathbb{RP}^3,\mathbb Z)=\mathbb Z,$ $H^2(\mathbb{RP}^3,\mathbb Z)=\mathbb Z/2\mathbb Z$ and other cohomology groups of $\mathbb{RP}^3$ are zero. I know that the fiber $F$ is a compact connected orientable manifold but I don't know more about it. I have no information about the total space $E$ and I want to find its cohomology. The $E_2$ page of the corresponding Serre spectral sequence, from $E^{s,t}_2=H^{s}(\mathbb{RP}^3,H^t(F)),$ is as below. $$\begin{matrix} \mathbb Z & 0 & \mathbb Z/2\mathbb Z & \mathbb Z \\ (\mathbb Z/2\mathbb Z)^2 & (\mathbb Z/2\mathbb Z)^2 & (\mathbb Z/2\mathbb Z)^2 & (\mathbb Z/2\mathbb Z)^2 \\ 0 & 0 & 0 & 0 \\ \mathbb Z & 0 & \mathbb Z/2\mathbb Z & \mathbb Z\end{matrix}$$ The only non-trivial differential is $d_2:E^{0,3}_2\rightarrow E_2^{2,2}$ and since it is a homomorphism from integers to $(\mathbb Z/2\mathbb Z)^2$ it cannot be surjective, so $d_2$ survives. And so by degree reasons $E_\infty$ degenerates to $E_3$ where every term in $E_3$ is same as to that of $E_2$ except at the index $(2,2)$ which I don't really know what it is, $E_3^{2,2}$ is either $\mathbb Z/2\mathbb Z$ or $(\mathbb Z/2\mathbb Z)^2$. I'm having difficulty calculating $E_3^{2,2}$. Since it is a Serre spectral sequence we say $E_2^{s,t}$ abuts to $H^{s+t}(E,\mathbb Z)$ but I couldn't find how to extract $H^{s+t}(E,\mathbb Z)$ from $E_\infty,$ is it generally the direct sum of terms $E_3^{i,s+t-i}=E_\infty^{i,s+t-i}$, if not, how to proceed?