The moment the stick comes to rest at θ=62.1° from horizontal.
Angular acceleration = (net torque) / (moment of inertia)
α = τ/I
We have to add up the torques due to the bugs and the stick; and add up the moments of inertia due to all three also.
Let L be the stick's length and let m be the stick's mass (so "2.75m" is each bug's mass). And let's say the "lower" ladybug is on the left. Then the lower ladybug exerts this much torque:
τ_lowerbug = −(2/5)L(2.75mg)cosθ (negative because I am (arbitrarily) choosing counter-clockwise as the negative angular direction).
The upper ladybug exerts this much torque:
τ_upperbug = +(3/5)L(2.75mg)cosθ
The weight of the stick can be assumed to act through its center, which is 1/10 of the way from the fulcrum. So the stick exerts this much torque:
τ_stick = +(1/10)L(mg)cosθ
The net torque is thus:
τ_net = τ_lowerbug + τ_upperbug + τ_stick
= −(2/5)L(2.75mg)cosθ + (3/5)L(2.75mg)cosθ + (1/10)L(mg)cosθ
= (2.75(3/5−2/5)+1/10)(mgL)cosθ
Now for the moments of inertia. The bugs can be considered point masses of "2.75m" each. So for each of them you can use the simple formula: I=mass×R²:
I_lowerbug = (2.75m)((2/5)L)² = (2.75m)(4/25)L²
I_upperbug = (2.75m)((3/5)L)² = (2.75m)(9/25)L²
For the stick, we can use the parallel axis theorem. This says, when rotating something about an axis offset a distance "R" from its center of mass, the moment of inertia is:
I = I_cm + mR²
We know that for a stick about its center of mass, I_cm is (1/12)mL² (see many sources). And in this problem we know that it's offset by R=(1/10)L. So:
I_stick = (1/12)mL² + m((1/10)L)²
= (1/12)mL² + (1/100)mL²
= (7/75)mL²
So the total moment of inertia is:
I_total = I_lowerbug + I_upperbug + I_stick
= (2.75m)(4/25)L² + (2.75m)(9/25)L² + (7/75)mL²
= (2.75(4/25+9/25)+7/75)mL²
So that means the angular acceleration is:
α = τ_net/I_total
= ((2.75(3/5−2/5)+1/10)(mgL)cosθ)/((2.75(4...
The "m" cancels out. You're given "L" and "θ" and you know "g", so do the math (and don't forget to use consistent units).
