Solving BJT transistor voltages and currents
The first thing to note, of course, is that you only need one of \$V_E\$, \$V_C\$, or \$V_B\$ in order to get the other two. And if you have those, you can compute \$I_C\$ and \$I_B\$ (obviously, \$I_E\$ is a given value here.) So the entire problem turns on being able to solve for one of those.

I'll just completely ignore your specific values and treat this more generally. I'm also just going to pick \$V_C\$ at random, base everything on that value, and see where that takes us. (It doesn't really matter which one I pick.) Finally, this is an NPN so I will consider \$I_E\$ as positive if exiting the emitter, \$I_C\$ as positive if entering the collector, and \$I_B\$ as positive if entering the base. This means that I can rely upon the usual \$I_E=I_C+I_B\$.

It's pretty easy to see that \$V_C=V_2-R_2\cdot I_C\$. But \$I_C\$ isn't known and it looks difficult to just guess at. Still, you also know that \$I_C=I_E-I_B\$, so that means we could try out this replacement and find \$V_C=V_2-R_2\cdot \left(I_E-I_B\right)\$. We are given \$I_E\$ and we can easily state that \$I_B=\frac{0\:\textrm{V}-V_B}{R_1}\$. So let's stuff all that together and see what this makes for us:

$$V_C=V_2-R_2\cdot \left(I_E-\frac{0\:\textrm{V}-V_B}{R_1}\right)$$

Hmm. We can replace \$V_B\$ by using \$V_C\$, right? We do know that \$V_B=V_E+700\:\textrm{mV}\$ and that \$V_C=V_E+100\:\textrm{mV}\$. So let's do that:

$$V_C=V_2-R_2\cdot \left(I_E-\frac{0\:\textrm{V}-\left(V_C+600\:\textrm{mV}\right)}{R_1}\right)$$

Think you can solve that for \$V_C\$? If so, do you think that may help solve all of the rest, as well?