It looks as though you are trying to use KCL and that the middle node at the top of your circuit is being called \$V_1\$ in your first equation in your question. (Potentially confusing, as you also have a labeled voltage source on the left with the same name.)
If so, then it is really:
$$\frac{24\:\text{V}-V_1}{250\:\Omega}+\frac{60\cdot I_b-V_1}{150\:\Omega}=\frac{V_1}{50\:\Omega}$$
This is not too far away from what you wrote. But I think you missed something in that second term on the left side and, perhaps, also got a sign wrong.
The above KCL I just wrote should be correct, if I'm not mistaken.
Also, you know that \$I_b=\frac{24\:\text{V}-V_1}{250\:\Omega}\$. So all you need to do is substitute that into the above equation, for \$I_b\$, and solve for the only remaining variable, \$V_1\$.
A properly arranged circuit for LTspice may be drawn out like this:
The above will solve out in LTspice and provide exactly the same value that you should get using the substitution I suggested.