Step 1: Picture the rod.
A uniform rod of mass 20 kg and length $L = 1.6$ m is hinged at one end and held horizontal. When released, gravity makes it swing down. We want the angular acceleration right at the start.
Step 2: Recall the rotation rule.
For rotation, torque equals moment of inertia times angular acceleration: $\tau = I \alpha$. So $\alpha = \tau / I$.
Step 3: Find the torque from gravity.
Gravity acts at the rod's centre, a distance $L/2$ from the hinge. So the turning torque is $\tau = M g \times \dfrac{L}{2}$.
Step 4: Write the moment of inertia.
For a rod pivoted at one end, $I = \dfrac{M L^{2}}{3}$.
Step 5: Combine to find angular acceleration.
$\alpha = \dfrac{M g (L/2)}{M L^{2}/3} = \dfrac{3 g}{2 L}$. The mass cancels out, which is a neat check.
Step 6: Put in the length.
With $L = 1.6$ m: $\alpha = \dfrac{3 g}{2 \times 1.6} = \dfrac{3 g}{3.2} = \dfrac{15}{16} g$. \[ \boxed{\dfrac{15}{16} g} \]