The formula to calculate the moment of inertia $\left(I\right)$ of the cylinder about its radial axis is given by

$I=M{r}^2 ...\left(1\right)$

Also, the torque $\left(\tau \right)$ on the cylinder about the central axis because of the application of the external force is given by

$\tau = Fr ...\left(2\right)$

Also, the torque can be expressed as

$\tau =I\alpha ...\left(3\right)$

Substitute the expressions from equation (1) and (2) into equation (3) and simplify to obtain the angular acceleration.

$\begin{array}{rcl}Fr& =& M{r}^2\alpha \\ & \Rightarrow & \alpha = \frac{F}{Mr} ...\left(4\right)\end{array}$

Substitute the values of the known parameters into equation (4) to calculate the required angular acceleration.

$\begin{array}{rcl}\alpha & =& \frac{52.5 \mathrm{N}}{5 \mathrm{kg}\times 0.70 \mathrm{m}}\\ & =& 15 \mathrm{rad} {\mathrm{s}}^{-2}\end{array}$