Answer
$a = 1.6~m/s^2$
Work Step by Step
We can assume that the frictional force $F_f$ is exerted in the +x direction.
We can use Newton's Second Law to find an expression for $F_f$:
$F_{app}+F_f = Ma$
$F_f = Ma-F_{app}$
We can consider the torque about the center to find the acceleration:
$\tau = I~\alpha$
$R~F_{app}-R~F_f = \frac{1}{2}MR^2~(\frac{a}{R})$
$F_{app}-F_f = \frac{1}{2}Ma$
$F_{app}-(Ma-F_{app}) = \frac{1}{2}Ma$
$2F_{app} = Ma+\frac{1}{2}Ma$
$2F_{app} = \frac{3}{2}Ma$
$a = \frac{4~F_{app}}{3M}$
$a = \frac{(4)(12~N)}{(3)(10~kg)}$
$a = 1.6~m/s^2$
