The source will have some internal voltage, so

\begin{aligned} -4+\frac{V_3}{3}+\frac{V_3-\left(-4 V_3\right)}{2  \Omega} & =0 \\ -4+\frac{V_3}{3}+\frac{5 V_3}{2} & =0 \\ \Rightarrow V_3 & =\frac{24}{17} \text { Volts } \\ \Rightarrow I=\frac{V_3+4 V_3}{2}=\frac{5}{2} \times \frac{24}{17} & =\frac{60}{17}  A \end{aligned}

Note: The voltage across ideal current source can be any value and it is decided by the other elements from the network.