Moment of inertia about an axis AB for a rod of mass 40 kg and...

Moment of inertia of the rod about axis $AB$ passing through one of its ends:
$I_{\text{rod}} = \dfrac{ML^2}{3} = \dfrac{40 \times (3)^2}{3} = \dfrac{40 \times 9}{3} = 120\text{ kg}\cdot\text{m}^2$

Moment of inertia of the solid sphere about the same axis $AB$, using the parallel axis theorem $I = I_{\text{cm}} + md^2$:
Mass $m = 10\text{ kg}$, radius $R$, distance from the centre of the sphere to axis $AB$ is $d = 3\text{ m}$.

$I_{\text{sphere}} = \dfrac{2}{5}mR^2 + md^2 = \dfrac{2}{5}(10)R^2 + 10(3)^2 = 4R^2 + 90$

Equating the two moments of inertia:
$120 = 4R^2 + 90$
$4R^2 = 30$
$R^2 = \dfrac{15}{2}$
$R = \sqrt{\dfrac{15}{2}}$

Comparing with $R = \sqrt{\dfrac{\alpha}{2}}$:
$\alpha = 15$

Hence, $\alpha = 15$.