The torque equations can be used to get expressions for F_{front} and F_{rear} in terms of m and α. Use your diagrams for the wheels to help set up these equations.

Umm, guys, i just did this using:
2F_{front} - 2F_{rear} = a(M + 4m),
τ - F_{front}R = Iα,
F_{rear}R = Iα,
but it gave me ##a = \frac {2τ} {R(M+2m)}## instead of ##a = \frac {2τ} {R(M+6m)}##

Can the OP or someone else explain how that equation was simplified to get a factor of 3 in the denominator for "m"?
The original equation involves a term with rotational kinetic energy (1/2Iω^{2}).

Was that "I" for the object's inertia converted to some kind of equivalent mass?