Apply Newton’s second law to the three objects assuming that the acceleration of the $2m$ mass is equal in magnitude and opposite in direction to that of the $4m$ mass.
\begin{align} -2ma =& T_1 - 2 mg \\ 4ma =& T_2 - 4mg \\ I \alpha =& T_1R - T_2R \end{align} The last equation needs help by expressing I and α in terms of $m$ and $a$, then simplifying. \begin{align} \frac{1}{2} (2mR^2 )\frac{ a}{R} =&( T_1 - T_2 )R\\ ma =&( T_1 - T_2 ) \end{align} We have 3 equations and 3 unknowns. Eliminate the $T$’s and solve for $a$: \begin{align} -6ma =& (T_1 - T_2) + 2mg \\ -6ma =& ma + 2mg \end{align} $$ a = -\frac{2}{7} g$$